81279 + Interview Questions in Mathematics QA in Mathematics Page 13 InterviewSolution

601.	142. From a tower 125 metres high, theangles of depression of two ob-jects, which are in horizontal linethrough the base of the tower, are45° and 30° and they are on thesame side of the tower. The dis-tance (in metres) between theobjects is(A) 125V3(B) 125(3-1)(C) 125/(3-1)(D) 125(V3 + 1)
Answer» AB = TOWERIn triangle ABC tan 45 = AB/BC1 = AB/AC = AB:BC = 1:1 In triangle ABD tan 30 = AB/BD= AB:BD = 1: root(3) CD = BD - BC = (root(3) - 1) units AB = 1 unit = 125 metres CD = (root(3) - 1) units = 125(root(3) - 1) metres (B) is correct option

601.

142. From a tower 125 metres high, theangles of depression of two ob-jects, which are in horizontal linethrough the base of the tower, are45° and 30° and they are on thesame side of the tower. The dis-tance (in metres) between theobjects is(A) 125V3(B) 125(3-1)(C) 125/(3-1)(D) 125(V3 + 1)

Answer»

AB = TOWERIn triangle ABC tan 45 = AB/BC1 = AB/AC = AB:BC = 1:1

In triangle ABD tan 30 = AB/BD= AB:BD = 1: root(3)

CD = BD - BC = (root(3) - 1) units

AB = 1 unit = 125 metres

CD = (root(3) - 1) units = 125(root(3) - 1) metres

(B) is correct option

Discussion

602.	I. An agent charges 12% commission on thesales. What does he earn if the total salesamount to 36,000? What does the seller get?
Answer»

Discussion

603.	8 ^ { 255 } = ( 32 ) ^ { x }
Answer»

Discussion

604.	255 में बेचने से 15% की हानि होती है। मेज का क्रय ।२. एक मेज कोमूल्य क्या है?(a) 200(०) * 30(c) १ 275(८ ३ 300
Answer» 255=85%ya 100%=255/85cost price =300

Discussion

605.	Ltd.3. The following balances has been extracted from the trial of M's Runway shinePrepare a trading and profit and loss account and a balance sheet as on December 31,2005.AmountAmountAccount TitleAccount Title2,50,0004,500PurchasesOpening stockReturn inwardsCarriage inwardsCash in handCash at bankWagesPrinting and stationeryDiscountBad debtsInsuranceInvestmentDebtorsBills receivablePostage and TelegraphCommissionInterestRepairLighting chargesTelephone chargesCarriage outwardMotor car1,50,000 Sales50,000 Return outwards2,000 Interest received4,500 Discount received77,800 Creditors60,800 Bill payable2,400 Capital1,25,0006,0401,00,0004,5001,5002,50032,00053,00020,0004002001,00044050010040025,0004,89,4404,89,440Adjustments1. Further bad debts1,000. Discount on debtors 500 and make a provision ondebtors 5%.2. Interest received on investment 5%.3. wages and interest outstandingて100 and 200 respectively.4. Depreciation charged on motorcar 5% p.a.5. Closing stock 32,500.
Answer»

Discussion

606.	(i)Total sales for Ram are Rs.21000,Rs.41000, Rs.31000. Total sales forRahim areRs.25000, Rs.38000,Rs.29000.
Answer» Ram = 21000+ 41000+31000= 93000, Rahim= 25000+ 38000 + 29000 = 92000, total sales for Ram = 93000; total sales for Rahm =92000; RAM=21000+41000+31000=93000RAHIM=25000+38000+29000=92000Total sale for ram=93000Total sale for Rahim=92000 total sales of ram =93000total sales of rahim=92000 92000 is the right answer of question 92000 is the most correct answer

606.

(i)Total sales for Ram are Rs.21000,Rs.41000, Rs.31000. Total sales forRahim areRs.25000, Rs.38000,Rs.29000.

Answer»

Ram = 21000+ 41000+31000= 93000, Rahim= 25000+ 38000 + 29000 = 92000, total sales for Ram = 93000; total sales for Rahm =92000;

RAM=21000+41000+31000=93000RAHIM=25000+38000+29000=92000Total sale for ram=93000Total sale for Rahim=92000

total sales of ram =93000total sales of rahim=92000

92000 is the right answer of question

92000 is the most correct answer

Discussion

607.	Sana deposited * 4000 in a bank for 1 – years. Thebank gives a half-yearly compound interest at 10%per annum. Find the compound interest,
Answer» For compound interestAmount = P(1 + R/100)^n = 4000(1 + 10/100)^n = 4000(11/10)^3 = 4(11)^3 = 41331 = 5324 Therefore,Compound interest= A - P= 5324 - 4000= Rs 1324

607.

Sana deposited * 4000 in a bank for 1 – years. Thebank gives a half-yearly compound interest at 10%per annum. Find the compound interest,

Answer»

For compound interestAmount = P(1 + R/100)^n = 4000(1 + 10/100)^n = 4000(11/10)^3 = 4*(11)^3 = 4*1331 = 5324

Therefore,Compound interest= A - P= 5324 - 4000= Rs 1324

Discussion

608.	Received cash from Amitt 19,000 in full settlement of his account. Paidinsurance premium 1,500
Answer» here the journal entry would be credit - cash 19000debit - insurance expense - 1500.

608.

Received cash from Amitt 19,000 in full settlement of his account. Paidinsurance premium 1,500

Answer»

here the journal entry would be

credit - cash 19000debit - insurance expense - 1500.

Discussion

609.	1: Reena gets an amount of Rs 50,000 from a bank at a simple interest of 10% per annumfor 2 years. Find the following.a) Amount of interestb) Total amount to be returned to bank after 2 years.
Answer» a) 50000102/100 = 10000 rupees b) 50000+10000 = 60000 rupees

609.

1: Reena gets an amount of Rs 50,000 from a bank at a simple interest of 10% per annumfor 2 years. Find the following.a) Amount of interestb) Total amount to be returned to bank after 2 years.

Answer»

a) 50000*10*2/100

= 10000 rupees

b) 50000+10000

= 60000 rupees

Discussion

610.	leachex ond clecs modes ckopansuoes rake Sense
Answer» Letx= 0.9999..There is only one repeating digit so multiply by 10 we get10x= 9.9999..x= 0.9999..subtract x now we get9x= 9Divide by 9 we getx= 9/9 = 1

Discussion

611.	7. If O is a point in the exterior of ABC, show that2(OA +OB+OC) (AB + BC + CA).Hint. Jotn OA, OB, OCFrom AAOB, OA +OB> ABFrom △BOC, OB +OC > BC.From AAOC, OA +OC > CA
Answer»

Discussion

612.	(i) 2 -3 - 83-3 का एक गुणनखंड (x + 1) है ।
Answer» (x+1) is one factor of polynomial 2x^3 - 3x^2 - 8x - 3 So,We can write given polynomial as f(x) = 2x^3 + 2x^2 - 5x^2 - 5x - 3x - 3= 2x^2(x+1) - 5x(x+1) - 3(x+1)= (x+1)(2x^2 - 5x - 3)= (x+1)(2x^2 - 6x + x - 3)= (x+1)[2x(x - 3) + 1(x - 3)]= (x+1)(2x+1)(x-3) Therefore, other two factors are (2x+1) and (x-3)

612.

(i) 2 -3 - 83-3 का एक गुणनखंड (x + 1) है ।

Answer»

(x+1) is one factor of polynomial 2x^3 - 3x^2 - 8x - 3

So,We can write given polynomial as f(x)

= 2x^3 + 2x^2 - 5x^2 - 5x - 3x - 3= 2x^2(x+1) - 5x(x+1) - 3(x+1)= (x+1)(2x^2 - 5x - 3)= (x+1)(2x^2 - 6x + x - 3)= (x+1)[2x(x - 3) + 1(x - 3)]= (x+1)(2x+1)(x-3)

Therefore, other two factors are (2x+1) and (x-3)

Discussion

613.	65-[83-\{29-(17-9-3)\}]
Answer» 0 APPLY BODMAS RULE TO SOLVE THIS QUESTIONSO, ANSWER WILL BE 0

613.

65-[83-\{29-(17-9-3)\}]

Answer»

0

APPLY BODMAS RULE TO SOLVE THIS QUESTIONSO, ANSWER WILL BE 0

Discussion

614.	मुझे 70 रुपये पर 70 पैसे लाभ प्राप्त होता है। मेरालाभ प्रतिशत है-(1) 10% (2) 7%(4) 1%(5) इनमें से कोई नहीं [Income Tax, 1992]
Answer» cp =70₹profit = 70p= .70₹%profit = profit ×100/cp = .70×100/70= 70×100/70×100=1% 1% is correct answer in this questions profit=profitx100/cp=70×100/70×100=1°\° 1% is correct answer of this problem. 1% is the correct answer 1 ℅ is a correct answer 4 is a correct answer simple questioncp 70profit. 70p=.70rs. 70*100/701℅ correct answer is 1% 4is a correct answer

614.

मुझे 70 रुपये पर 70 पैसे लाभ प्राप्त होता है। मेरालाभ प्रतिशत है-(1) 10% (2) 7%(4) 1%(5) इनमें से कोई नहीं [Income Tax, 1992]

Answer»

cp =70₹profit = 70p= .70₹%profit = profit ×100/cp = .70×100/70= 70×100/70×100=1%

1% is correct answer in this questions

profit=profitx100/cp=70×100/70×100=1°\°

1% is correct answer of this problem.

1% is the correct answer

1 ℅ is a correct answer

4 is a correct answer

simple questioncp 70profit. 70p=.70rs. 70*100/701℅

correct answer is 1%

4is a correct answer

Discussion

615.	R-83-240Q7. Simplify <3%( )".
Answer» (-3) ⁴ ×(5/3) ⁴(5) ⁴625

Discussion

616.	ro is a point within ΔΑΒC, show that(i) AB+ AC> OB+ OCii) AB + BC CA >OA+ OB+ OC(iii) OA+OB+ OC (AB+ BC+ CA)2
Answer» can u give the photo I cannot see the half part of the and plz

Discussion

617.	O is a point within AABC, show that:(i) AB + AC > OB + OC(ii) AB + BC + CA > OA +OB + OC(i) 04 OB+OAB+BC+CA)
Answer»

Discussion

618.	fO is a point within AABC, show that:(i) AB + AC >OB+OC(ii) AB + BC + CA>OA + OB +OC() O4 OB+OC AB+ BC+CA)2
Answer»

Discussion

619.	1 In the Fig. 2, given below, OB is the perpendicular bisector of the line segment DE, FA LOB andFE intersects OB at the point C. Prove thatROA OB OC
Answer» OB is perpendicular bisector of line segment DE , FA perpendicular to OB and FE intersects OB at the point C as shown in figure . now, ∆OAF and ∆ODB ∠OAF = ∠OBD = 90° {because OB is perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF } ∠FOA = ∠DOB { common angle } from A - A similarity , ∆OAF ~ ∆ODB so, OA/OB = AF/DB = OF/OD -------(1) similarly, ∆AFC and ∆BEC ∠FCA = ∠BCE ∠FAC = ∠CBE = 90° from A - A similarity , ∆AFC ~ ∆BEC so, AF/BE = AC/CB = FC/CE -------(2) we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2) AF/DB = AC/CB =FC/CE -------(3) now, from equations (1) and (3), OA/OB = AC/CB = ( OC - OA)/(OB - OC) OA/OB = (OC - OA)/(OB - OC) OA(OB - OC) = OB(OC - OA) OA.OB - OA.OC = OB.OC - OB.OA 2OA.OB = OB.OC + OA.OC dividing by OA.OB.OC both sides, 2OA.OB/OA.OB.OC = OB.OC/OA.OB.OC + OA.OC/OA.OB.OC2/OC = 1/OA + 1/OB

619.

1 In the Fig. 2, given below, OB is the perpendicular bisector of the line segment DE, FA LOB andFE intersects OB at the point C. Prove thatROA OB OC

Answer»

OB is perpendicular bisector of line segment DE , FA perpendicular to OB and FE intersects OB at the point C as shown in figure .

now, ∆OAF and ∆ODB ∠OAF = ∠OBD = 90° {because OB is perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF } ∠FOA = ∠DOB { common angle } from A - A similarity , ∆OAF ~ ∆ODB so, OA/OB = AF/DB = OF/OD -------(1)

similarly, ∆AFC and ∆BEC ∠FCA = ∠BCE ∠FAC = ∠CBE = 90° from A - A similarity , ∆AFC ~ ∆BEC so, AF/BE = AC/CB = FC/CE -------(2)

we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2) AF/DB = AC/CB =FC/CE -------(3)

now, from equations (1) and (3), OA/OB = AC/CB = ( OC - OA)/(OB - OC) OA/OB = (OC - OA)/(OB - OC) OA(OB - OC) = OB(OC - OA) OA.OB - OA.OC = OB.OC - OB.OA 2OA.OB = OB.OC + OA.OC dividing by OA.OB.OC both sides, 2OA.OB/OA.OB.OC = OB.OC/OA.OB.OC + OA.OC/OA.OB.OC2/OC = 1/OA + 1/OB

Discussion

620.	27. In the Fig. 2, given below, OB is the perpendicular bisector of the line segment DE, FA 1 OB andFE intersects OB at the point C. Prove that 01+-OBOC
Answer» OB is perpendicular bisector of line segment DE , FA perpendicular to OB and FE intersects OB at the point C as shown in figure . now, ∆OAF and ∆ODB ∠OAF = ∠OBD = 90° {because OB is perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF } ∠FOA = ∠DOB { common angle } from A - A similarity , ∆OAF ~ ∆ODB so, OA/OB = AF/DB = OF/OD -------(1) similarly, ∆AFC and ∆BEC ∠FCA = ∠BCE ∠FAC = ∠CBE = 90° from A - A similarity , ∆AFC ~ ∆BEC so, AF/BE = AC/CB = FC/CE -------(2) we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2) AF/DB = AC/CB =FC/CE -------(3) now, from equations (1) and (3), OA/OB = AC/CB = ( OC - OA)/(OB - OC) OA/OB = (OC - OA)/(OB - OC) OA(OB - OC) = OB(OC - OA) OA.OB - OA.OC = OB.OC - OB.OA 2OA.OB = OB.OC + OA.OC dividing by OA.OB.OC both sides, 2OA.OB/OA.OB.OC = OB.OC/OA.OB.OC + OA.OC/OA.OB.OC2/OC = 1/OA + 1/OB JAG's she did say she a single day

620.

27. In the Fig. 2, given below, OB is the perpendicular bisector of the line segment DE, FA 1 OB andFE intersects OB at the point C. Prove that 01+-OBOC

Answer»

OB is perpendicular bisector of line segment DE , FA perpendicular to OB and FE intersects OB at the point C as shown in figure .

now, ∆OAF and ∆ODB ∠OAF = ∠OBD = 90° {because OB is perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF } ∠FOA = ∠DOB { common angle } from A - A similarity , ∆OAF ~ ∆ODB so, OA/OB = AF/DB = OF/OD -------(1)

similarly, ∆AFC and ∆BEC ∠FCA = ∠BCE ∠FAC = ∠CBE = 90° from A - A similarity , ∆AFC ~ ∆BEC so, AF/BE = AC/CB = FC/CE -------(2)

we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2) AF/DB = AC/CB =FC/CE -------(3)

now, from equations (1) and (3), OA/OB = AC/CB = ( OC - OA)/(OB - OC) OA/OB = (OC - OA)/(OB - OC) OA(OB - OC) = OB(OC - OA) OA.OB - OA.OC = OB.OC - OB.OA 2OA.OB = OB.OC + OA.OC dividing by OA.OB.OC both sides, 2OA.OB/OA.OB.OC = OB.OC/OA.OB.OC + OA.OC/OA.OB.OC2/OC = 1/OA + 1/OB

JAG's she did say she a single day

Discussion

621.	\| किसी / ABC में यदि तीनों माध्यिकाएँ 0 बिन्दु पर परस्परप्रतिच्छेद करती हैं, तो AB+B+AC का मान निम्नलिखित\|में से किसके बराबर है?\| (a) OAP+OB+OC? (b) 2(OA+OB+OCP)\| (c) 3(OA+OB+OC) (d) 2(OA+OB2+OC)
Answer» 2(OA^2+OB^2+OC^2) is a right answer

Discussion

622.	5.In Fig. AB > AC and OB and OC are the bisectors of ZB and /C respectively. Show thatOB > ОС.
Answer»

Discussion

623.	In the given figure, O is a point in the exterior ofShow that:triangle Auu(i) OA OB >AB(ii) OB + OC > BC(ii) OC + OA > CA1ul(iv) 2(0A + OB + OC)> (AB + BC+CA).
Answer»

Discussion

624.	12, Which will earn more interest and how much?i. Rs 6000 lent at 12% pa. compounded annually for 1 1/2 years.ii.Rs 6000 lent at 12% pa. compounded half yearly for 1 1/2 years
Answer» 1) A=6000(1+0.12)^1.5=6000(1.12)^1.5=7111.78so interest=1111.782) A=6000(1+0.12/2)^3=6000(1.06)^3=7146.1so interest=1146.1 so 2nd get more interest

624.

12, Which will earn more interest and how much?i. Rs 6000 lent at 12% pa. compounded annually for 1 1/2 years.ii.Rs 6000 lent at 12% pa. compounded half yearly for 1 1/2 years

Answer»

1) A=6000(1+0.12)^1.5=6000(1.12)^1.5=7111.78so interest=1111.782) A=6000(1+0.12/2)^3=6000(1.06)^3=7146.1so interest=1146.1

so 2nd get more interest

Discussion

625.	compounded half yearlyhe amount and the compound Interest oncompounded semi-annually12800 for 1 year at7per annumfor 2 years at 10% per anni
Answer»

Discussion

626.	8.Find the amount and the compound interest ont10.000 for 1-years at 10%2annum, compounded half yearly. Would this interest be more than the interest hewould get if it was compounded annually?
Answer»

Discussion

627.	. Mukesh borrowed 75000 from aIf the rate of interest is 12% per annum, findthe amount he would be paying after 1; years if the interest is(i) compounded annually(ii) compounded half-yearly
Answer» If Mukesh borrowed 75,000 from a bank and the rate of interest is 12% per annum, Then , 12% of 75,000=> ₹(12 × 75,000 / 100)=> ₹(12 × 750)=> ₹9,000 Total interest per annum = ₹(75,000+9,000)= ₹ 84,000 In one and half years = ₹ (84,000 + 42,000)= ₹ 1,26,000 Compound interest = 12% of 84000= ₹ 10080 Compounding annually = ₹ (10080 + 84000)= ₹ 94080

627.

. Mukesh borrowed 75000 from aIf the rate of interest is 12% per annum, findthe amount he would be paying after 1; years if the interest is(i) compounded annually(ii) compounded half-yearly

Answer»

If Mukesh borrowed 75,000 from a bank and the rate of interest is 12% per annum,

Then , 12% of 75,000=> ₹(12 × 75,000 / 100)=> ₹(12 × 750)=> ₹9,000

Total interest per annum = ₹(75,000+9,000)= ₹ 84,000

In one and half years = ₹ (84,000 + 42,000)= ₹ 1,26,000

Compound interest = 12% of 84000= ₹ 10080

Compounding annually = ₹ (10080 + 84000)= ₹ 94080

Discussion

628.	- Mioprepon simple cash book in the book2015- Oct -opening cash Balance 5000good panchese 1200Salary paid 1.100commission Received 300cash received from Geite 500.Good sales to kanan 280paid Rent to Abhi 100
Answer» 5000-1200=3800-1120=2680+300=2980+500=3480+280=3760-100=3660.

Discussion

629.	At what rate per cent per annum will 3600 amount to e4734 in 37 years?2
Answer»

Discussion

630.	ISE1Beof formula for CIFind the amount and compound interest on(i) Rs 12,000 in 2 years at 8% per annum
Answer»

Discussion

631.	has a thickness 5em. 4fnd the amoun o bead ice semaineal atteroutside temperature Δ45e and kesv thermoceliaろ
Answer»

Discussion

632.	te following questions:If thee amount becomes Rs. y for any principal having the rate of simple interest of t/e perChoose tle1x6-6for x years, then the principal will beannum(a) Rs. xyz(c) Rs100+xz(b) Rs. XZ100y100z(d) Rs. 100+K
Answer» option C is the correct answer

Discussion

633.	MATHEMATICSat simple interest andper annum,3. Fabinaborrows? 12,500 at 12% per annum for 3 yearsRadha borrows the same amount for tcompounded annually. Who pays more interestthe same time period at 10%e inand by how much?nnum simple interest for 2 years
Answer»

Discussion

634.	A shopkeeper sells an article for 248.50 atterallowing a discount of 10%. Find the price atwhich the article is marked.tinl for450. He
Answer»

Discussion

635.	50 men for 28 days. But, 700 men atter.In an army camp, there were provisions for 5the camp. How long did the provisions last?
Answer»

Discussion

636.	A shopkeeper incurred a loan of Rs.5,00,000/-at simple interestfor a period of five years. Atter 5 years he repaidRs.6,60,500/-What is the rate of interest?
Answer»

Discussion

637.	Application :3. Shrabani has got an interest of 45 by depositing some money in thebank. If the simple interest of the bank is 5% per annum, then let us write, by calculating,the amount of money deposited by shraboni in the bank
Answer»

Discussion

638.	) atter 1 years?Shanu took a loan of 60,000 from a bank. If the rate of interest is 8% per annum, findthe amount he would be paying after 1 years if the interest is(a) compounded annually7.2(b)compounded half yearly
Answer»

Discussion

639.	If the area of a triangle is 0 square units then the vertices of atriangle are(Fill in the blank) 1
Answer» If the area of triangle is 0 sq units then the vertices of triangle are collinear i.e.x1(y2-y3) + x2(y3-y1) + x3(y1-y2) =0

639.

If the area of a triangle is 0 square units then the vertices of atriangle are(Fill in the blank) 1

Answer»

If the area of triangle is 0 sq units then the vertices of triangle are collinear

i.e.x1(y2-y3) + x2(y3-y1) + x3(y1-y2) =0

Discussion

640.	EXAMPLE 12 The area of the triangle formed by the coordinate axes and a line issquare units and the length of its hypotenuse is 5 units. Find the equationof the line.
Answer»

Discussion

641.	If Suresh and Mahesh can do a work in 12 days, Mahesh and Ramesh in 15 days, Suresh andRamesh in 20 days, how long will each Suresh, Mahesh and Ramesh take to finish the work9.
Answer» Mahesh and suresh can do a work in =12days mahesh and suresh 1 day work = 1/12mahesh and ramesh can do work in = 15 daystheir one day work= 1/15 suresh and ramesh can do work in = 20 days their one day work = =1/20 so, the total time all of them take to complete the work is = 1/12+1/15+1/20 = 12/60 hit like if you find it useful full solution

641.

If Suresh and Mahesh can do a work in 12 days, Mahesh and Ramesh in 15 days, Suresh andRamesh in 20 days, how long will each Suresh, Mahesh and Ramesh take to finish the work9.

Answer»

Mahesh and suresh can do a work in =12days mahesh and suresh 1 day work = 1/12mahesh and ramesh can do work in = 15 daystheir one day work= 1/15 suresh and ramesh can do work in = 20 days their one day work = =1/20

so, the total time all of them take to complete the work is = 1/12+1/15+1/20 = 12/60

hit like if you find it useful

full solution

Discussion

642.	find the ratio of to millimetre in 1m?
Answer» 1m=100cm1cm=10mmSo,1m=100(10)mm 1m=1000mmOn equating we get,1000mm=100cmmm/cm=(100/1000)mm/cm=1/10=0.1 thank you to give me the answercan you tell me this question answer?the boys and girls in a school are in the ratio 8 :3 if the number of girls is 405 how many boys are there in a school?

642.

find the ratio of to millimetre in 1m?

Answer»

1m=100cm1cm=10mmSo,1m=100(10)mm

1m=1000mmOn equating we get,1000mm=100cmmm/cm=(100/1000)mm/cm=1/10=0.1

thank you to give me the answercan you tell me this question answer?the boys and girls in a school are in the ratio 8 :3 if the number of girls is 405 how many boys are there in a school?

Discussion

643.	The weight of Ramesh is 5 kg less than twice theweight of Mahesh. If their total weight is 115 kgfind the weight of Ramesh.
Answer» Let weight of Mahesh be x So, weight of Ramesh is 2x - 5 According to question, x + 2x - 5 = 115 3x = 120 x = 40 So, weight of Ramesh is 2×40-5= 80-5= 75 Thanks

643.

The weight of Ramesh is 5 kg less than twice theweight of Mahesh. If their total weight is 115 kgfind the weight of Ramesh.

Answer»

Let weight of Mahesh be x

So, weight of Ramesh is 2x - 5

According to question,

x + 2x - 5 = 115

3x = 120

x = 40

So, weight of Ramesh is 2×40-5= 80-5= 75

Thanks

Discussion

644.	Anita is twice asold as Mahesh. Ten years ago, Anita was seven times as old as Mahesh. Find their present ages.
Answer» Let the present ages of Manish and Anita be m and a respectively. a = 2m (a - 10) = 7(m - 10) 2m - 10 = 7m - 70 5m = 60 m = 12 a = 24Anita's present age = 24 yearsManish's present age = 12 years

Discussion

645.	A man completes 30 km of a journey at 6 km/hr and the remaining 40 km in 5 hours. Find hisaverage speed for whole journey.(a) 6km/h(b) 7 km/h- km/h(d) 8 km/h
Answer» option a is right answer according to your question the answer is 7Km/hr

645.

A man completes 30 km of a journey at 6 km/hr and the remaining 40 km in 5 hours. Find hisaverage speed for whole journey.(a) 6km/h(b) 7 km/h- km/h(d) 8 km/h

Answer»

option a is right answer according to your question

the answer is 7Km/hr

Discussion

646.	The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:
Answer»

Discussion

647.	The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:4.
Answer»

Discussion

648.	lum HumBer of seats?The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:l.Length (in mm)118-126127-135136-144145-153154-162163-171172-180Number of leaves12NOTE5 PRO42
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649.	find the base height 31.4 millimetre area of triangle 1256 millimetre square units
Answer» Area of a triangle = 1/2 × base × hight1256= 1/2 × base × 31.4base × 31.4 = 1256×2base = 80

Discussion

650.	The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:4.Number of leavesLength (in mm)118-126127-135136-144145-153154-162163-171172-18012420 Draw a histogram to represent the given data. [Hint: First make the class intervalscontinuous)(i) Is there any other suitable graphical representation for the same data?(ii) Is it correct to conclude that the maximum number of leaves are 153 mm long?Why?
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The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:

The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:4.

lum HumBer of seats?The length of 40 leaves of a plant are measured correct to one millimetre, and theobtained data is represented in the following table:l.Length (in mm)118-126127-135136-144145-153154-162163-171172-180Number of leaves12NOTE5 PRO42

find the base height 31.4 millimetre area of triangle 1256 millimetre square units

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R-83-240Q7. Simplify <3%( )".

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