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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 80401. |
3) Find the product of 3 and 2 and then take away 5 |
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Answer» According to question(3×2) - 56 - 5 = 1 thq u |
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| 80402. |
CRCe seras and SyFind the sum of 6 and 4 and then take away 2 |
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Answer» sum implies addition. take away isvsubtraction so 6+4-2 = 8 is the answer thank you very much I can send 8 answers |
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| 80403. |
result now obtained is 3 tnies ie l2. Apasilive nĂšnberis 5 times another number.If21 is added to both the numbersitive nidaberis 5 times another number. If21 is added to both the numbers,then one of the new numbers becomes twice the other new number. What are thenumbers?t number Ä°s O. When we interchange the digits, it is |
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Answer» Let one number = nThen second number = 5n If 21 added to both numbers Then according to the given condition(5n + 21) = 2(n + 21)5n + 21 = 2n + 425n - 2n = 42 - 213n = 21n = 21/3 = 7 Numbers are n = 21 and 5n = 5*21 =105 |
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| 80404. |
If you have a bowl with six apples and youtake away four, how many do you have |
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Answer» 6 |
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| 80405. |
Write the equation for the following statement: If you take away 6 from 6 times of yyou get 66. |
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Answer» The equation is66=6y-6 According to the problem, 6y- 6 = 66 |
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| 80406. |
Find three consecutive even number whose sum is 96.The sum of three consecutive integers is 39. Find all three integers |
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Answer» three consecutive even numbers are x-2,x,x+2sl x-2+x+x+2=96so 3x=96 so x=32so numbers are 30,32,34 three consecutive integers are x-1,x,x+1so x-1+x+x+1=39so 3x=39 so x=13so numbers are 12,13,14 |
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| 80407. |
2. One number is 3 times another number. If the smaller number is subtracted fromresult is 20 more than two-thirds the larger number. Find the numbers |
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| 80408. |
2 A positive number is 3 times another number. If 3 is added to the smaller number the differencebetween new number and greater becomes 31. What are the numbers? |
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| 80409. |
Determine the AP whose third term ıs l 6 and the 7th term exceeds te fit termi |
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Answer» Let a be the First term, a3 be the third term, a5 be the 5th term and a7 be the 7th term a3 = 16 a7 = a5 + 12 ............ (1) Let the common difference be "d" Common difference is equal in AP So, a7 = a5 + d + d = a5 + 2d............(2) From Equation (1) & (2) a5 + 12 = a5 + 2d 2d = 12 d = 6 From Given, we get that a3 = 16 a3 = a + 2d = 16 a + ( 2× 6 ) = 16 [ We know that d = 6 ] a + 12 = 16 a = 4 So first term is 4 .... We can find AP by adding d continuously So,AP is 4, 10, 16, 22, 28..... |
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| 80410. |
17. Determine an AP whose third term is 9 and when fifth term is subtracted from sth term, we get 6.OR |
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Answer» Let first term is a and common difference is d and nth term is a+(n-1)d9=a+(3-1)da+2d=9a+7d-a-4d=63d=6; d=2a=9-4=5so, AP is 5,7,9,11,13... |
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| 80411. |
lo.Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. |
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| 80412. |
5. In Fig. 6.32, if AB I| CD, Z APQ 50° andZ PRD1270, find xand y.50°1270R D |
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Answer» y + 50° = 127 ( alternative angle) y = 127° - 50° = 77° x = 50° ( Alternate angle) |
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| 80413. |
Fig. 6.32, if AB Il CD, Z APQ/ Z PRD 127°, find x and y.50° and |
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| 80414. |
1 ^ { 2 } + ( 1 ^ { 2 } + 2 ^ { 2 } ) + ( 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } ) + |
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| 80415. |
d the sum of the following algebrp r-6 p^{2} r, 3 p^{2} r-2 p r, 2 r p-p^{2} r |
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| 80416. |
r ^ { 3 } - 4 + 3 r ^ { 2 } , \text { take away } 5 r ^ { 2 } - 3 r ^ { 3 } + r - 7 |
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| 80417. |
write down three consecutive whole number just preceding 729001 |
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Answer» 7510001-3= 7509998. Therefore, 7510000,7509999 and 7509998 are3 consecutive whole numbers just preceding 7510001.Jul 6, 2018 |
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| 80418. |
4 The difference between the squares of two consecutive whole numbers is 31.numbers |
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| 80419. |
One number is 4 less than 3 times another. If their sumis increased by 5, the result is 25. Find the numbers. |
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| 80420. |
Determine the AP whose third term is 16 and the 7th termexceedsthe5thtermbyl2 |
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| 80421. |
.13) Find the sum of first 21 terms of an AP, whose 2nd term is8 and 4th term is 14 |
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| 80422. |
In the given figure, if AB II CD, find angles x and y.P10915. |
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Answer» x=136y=224is the best answer |
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| 80423. |
in the figure, AB //CD APQ= 50 and PRD= 127" Find angles x & y. |
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| 80424. |
In the given figure, if AB I| CD, find angles x and y |
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| 80425. |
1. Determine which of the following polynomials has (r+ 1) a factor:Gi) ++x+ 1(iv) r--(2+2)+2 |
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| 80426. |
a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1} |
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| 80427. |
\operatorname { s } \text { of } \left[ \begin{array} { r r r } { 7 } & { 2 } & { 2 } \\ { - 6 } & { - 1 } & { 2 } \\ { 6 } & { 2 } & { - 1 } \end{array} \right] |
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| 80428. |
Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers. |
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| 80429. |
1. Anis a statement that two quantities are not equal.2. The set of all whole numbers and their opposites areNumbers less3. Numbers greater than 0 arethan O are |
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Answer» 1. Inequality 2. Integers 3. Positive integers, negative integers |
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| 80430. |
Determine the A.P. whose 3Td term is 5 and the 7h term is 9. |
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Answer» First term be a Common difference be dnth term=a+(n-1)d 3rd term=a+2d=57th term=a+8d=9 Subtract 6d=9-5=4d=2/3a=5-2(2/3)=5-4/3=11/3 AP is 11/3, 13/3,15/3... |
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| 80431. |
Q. 16) In the figure, AB // CD LAPQ = 50° and <PRD = 127" Find angles x & y.50%t 2-127" / / .RD |
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Answer» x = 50° ( vertically opposite angle)y + 50° = 127° ( exterior angle theorem)y = 127° - 50°y = 77° |
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| 80432. |
9. Find the GP whose 4th and 7h terms are 1/18 &-1/486respectively |
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| 80433. |
5. If AB 11 CD,LAPO = 50, and LPRD127,ㄗfind x and y50027°c QA D |
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| 80434. |
SO-3016) in the figurend Her.APQ= 50 and LPRD-127"Find angles x &50127 |
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Answer» x = 50° ( Alternate angle)127° = y + 50°y = 127° - 50 ° ( Alternate angle)y = 87° |
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| 80435. |
1. (a+)+2) 93x-1(x-1) (x-2)(x-3)3、2x+3x +25.(x (x-2) (x-3)1-12(I-2x)(x-1)2 (x+2)· |
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Answer» 4) |
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| 80436. |
Fig. 6.31. In Fig. 6.32. if AB I CD, 4 APO 50° and A50°PRD 127, find x andy12R D |
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| 80437. |
\frac{(x+2)^{2}(x-1)^{2}}{(x-1)^{2} x^{2}} |
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| 80438. |
Whole Numbers3. The largest number of 6 digits which is exactly divisible by 16 is0094d) 999 |
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Answer» The largest six digit number is 999999. In order to find the largest six digit number divisible by 16 , divide 999999 by 16. You will get the remainder 15. Now subtract 15 from 999999. Hence, the largest six digit number divisible by 16=999999-15=999984 999984 is the correct answer of the given question largest 6 digit no. =999999if 999999/16 then remainder is 15 so 999999-15=999984 which is divided by 16 if u satisfied with the ans then like or cmmt ✌️ 999984 is the answer of the following 999984 is the correct answer |
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| 80439. |
verify rolle's theoram for the following functions:- 1.f(x)=(x-2)(x-3)(x-4) in [2,4] |
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| 80440. |
69x 10. Find the least number of five digits that is exactly divisible byP.T.O16, 18, 24 and 30. |
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| 80441. |
A playground has the shape of a rectangle, with two semi-circles on its smaller sidesas diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m,find the area of the playground. (Take π=22/7). |
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| 80442. |
(E)Nneuruc-a, then the value of (1-x)(1-)-2 isa-b b-cthen the value ofe 13. If xya +(A) abc(C) 1(E) of these(B) a?b'c2(D)-1 |
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| 80443. |
50127c oR D |
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Answer» if AB||CD so PQR = 50and PRQ+PRD = 180 PRQ= 180-127= 53therefore PQR+PRQ+RPQ= 180y= 180-53-50=77 |
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| 80444. |
Ju that the resulting expression has131() In the given figure ABCD is a rectangle. It consists of a circle and two semi-which are of radius 5 cm. Find the area of the shaded region. Give youthree significant figuresanswer correct to |
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Answer» Breadth= 5*2 = 10cmLength = 5*4 = 20 cmArea of shaded region = Area of rectangle - Area of two semicirles - Area of circle= 10*20 - 2*(22/7)*5*5= 200 - 157.14= 42.86 sq cm |
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| 80445. |
iv) 40, 36 and 1263. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.(v) 84, 90 and 120(Vi) 2ona and 13 m 20 cm broad. It is to |
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Answer» Prime factorization of,24=2³×315=3×536=2²×3² LCM=product of each prime factor of highest power LCM=2³×3²×5=360 Greatest six digit number=999999 Greatest six digit number exactly divisible by given numbers=999999-remainder when 999999 is divided by LCM of given numbers Greatest six digit number exactly divisible by given numbers=999999-279=999720 Hence greatest six digit number exactly divisible by 24 , 15 and 36 is 999720. |
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| 80446. |
According to Pythagorus Theoram (35 |
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Answer» Ans :- Option (a) is correct. |
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| 80447. |
A 50905. q127C gR D6 |
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Answer» Assuming given in question that AB||CD 50° + y = 127° ( Alternate angle) y = 127° - 50° = 77° x = 50° ( Alternate angle) |
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| 80448. |
18. For what value of 2 is the function defined byJax – 2x), if x 50f(x)=3| 4x+1, if x>0.continuous at x=0? What about continuity at x = 12 |
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| 80449. |
A paper is in the form of a rectangle PQRS in whichPQ = 10 cm and QR = 7 cm. Two semi-circular portionswith QR and PS as diameters are cut off (As shown infigure 11.22). Find the area of the remaining part. |
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| 80450. |
). A paper is in the form of a rectangle PQRS in whichPQ = 10 cm and QR = 7 cm. Two semi-circular portionswith QR and PS as diameters are cut off (As shown infigure 11.22). Find the area of the remaining part. |
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Answer» Area of remaining part = Area of rectangle-2 area of semicircle= lb-2πr^2/2= (10)(7) - 2((22/7) (3.5)^2/2)= 70-38.5= 31.5 cm^2 |
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