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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.
| 63101. |
DBP one boot of the equation8C?-48 HEO is 2tJz then find theother root? |
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| 63102. |
5 IA.8 and C ana 4h points on a lineB is eetveon A and C than pmoue Ac,8c- na0ns |
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| 63103. |
16. A number decreased by 27% gives 87. The number is |
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| 63104. |
9. Find the median of the given data:6 7 52 10 93f 9 12 8 13 11 14 7 |
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Answer» so... median is 6 |
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| 63105. |
6) 7+ 10 1/2+14+. . +84 |
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Answer» thank you.. |
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| 63106. |
.A plot is 120m long and 105m broad. A path 10m wide is to be built around it. Find(1)the area of the path(ii)the area of the plot in hectare |
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Answer» area of plot without path = 12600 sq m width of path=10 m area of plot with path=(120+10+10)*(105+10+10) =17500 sq m area of path = 17500-12600 =4900 sq m |
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| 63107. |
A plot is 120m long and 105m broad. A path 10m wide is to be built around it. Find(i)3.the area of the path(i) the area of the plot in hectare. |
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Answer» Answer: 4900 sq m 49 hectares Step-by-step explanation: area of plot without path = 12600 sq m width of path=10 m area of plot with path=(120+10+10)*(105+10+10) =17500 sq m area of path = 17500-12600 =4900 sq m hit like if you find it useful |
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| 63108. |
3. A plot i(i)$ 120m long and 105m broad. A path 10m wide is to be built around it. Findthe area of the pathii) the area of the plot in hectare |
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| 63109. |
a A class room is 7m long, 6.5m wide and 4m high. It has one door 3m X 1.4m and threewindows measuring 2m x 1m. The interior walls are to be painted. Find the cost of painting atthe rate of Rs 3.50 per m2. |
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Answer» Area of four walls=(2h(l+b)) =(2x4(7+6.5)) =(8x13. 5) =108m square area of one window =3mx1. 4m =4.2m square area of three window =3(2mx1m) =6m square total area of windows =(4.2+6)m square =10.2m square Area for whitewashing =(108-10.2)m square =97.8m square cost of 1 m square white washing = 3.50rupee cost of 97.8m square white washing=3.50x97.8 =342.3 rupees |
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| 63110. |
rectangular lawn 80 m × 60 m has two roads each 10 m wide running in the middle of it, one parallelto the length and the other parallel to the breadth. Find the cost of gravelling them at 1.20 per sq metre |
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Answer» I I didn't understand this question |
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| 63111. |
28. A class room is 7m long, 6.5m wide and 4m high. It has one door 3m X 1.4m and threcwindows measuring 2m x 1m. The interior walls are to be painted. Find the cost of painting atthe rate of Rs 3.50 per m2. |
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Answer» Area of four walls=(2h(l+b))=(2x4(7+6.5))=(8x13. 5)=108m squarearea of one window =3mx1. 4m=4.2m squarearea of three window =3(2mx1m)=6m squaretotal area of windows =(4.2+6)m square=10.2m squareArea for whitewashing =(108-10.2)m square=97.8m square cost of 1 m square white washing = 3.50rupeecost of 97.8m square white washing=3.50x97.8=342.3 rupees |
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| 63112. |
7. A rectangular lawn 80 m × 60 m has two roads each 10 m wide running in the middle of it, one paral-lel to the length and the other parallel to the breadth. Find the cost of gravelling them at R 1.20 persq metre. |
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| 63113. |
find the expenditure of putting a barbed wire around a 4 hectare square field, if it costs Rs 45 per metre |
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Answer» Area of square shaped farm = 4 hectares = 10,000 m² [1 hectare = 10,000 sq m] Hence area of farm = 40,000 square meters. Area of square = (side)² (side)² = 40,000 side = √40000 side = 200 m Perimeter of square= 4(side) 4 × 200m 800 Given cost of fencing is Rs. 45 per meter Hence total cost for fencing 800 m= 45 × 800 = Rs. 36,000 thanks |
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| 63114. |
a rectangular field is 81 metre long and 49 metre wide a square has the same area as that of rectangular field find the side of the square |
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| 63115. |
\frac { a } { a x - 1 } + \frac { b } { b x - 1 } = a + b , a + b \neq 0 , a b \neq 0 |
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Answer» how did u do the 4th step hey? our? |
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| 63116. |
EXERCISE -8Crsmumber is 8 moret number. The sum of the two numbers is 70. Find the numb |
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Answer» suppose numbers are x and 70-x so x = 70-x+8 2x = 78 so x=39 so both the numbers 39 and 31. |
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| 63117. |
pe-21, A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the numb |
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| 63118. |
cost ofPLE 22 Two cross roads, each of widthof a rectangular park of length 70 m ands roads, each of width 5 m, run at right angles through thelar park of length 70 m and breadth 45 m and parallel to:sides. Find the area of the roads. Also find the cost of constructiroads at the rate of 105 per m².the cost of constructing the |
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Answer» LetABCDbe the rectangular park thenEFGHandIJKLthe two rectangular roads with width 5 m.Length of the rectangular parkAD= 70 cm Breadth of the rectangular parkCD= 45 m∴ Area of the rectangular park = Length x Breadth = 70 m x 45 m = 3150 m2Area of the roadEFGH= 70 m x 5 m = 350 m2Area of the roadJKIL= 45 m x 5 m = 225 m2Clearly area ofMNOPis common to the two roads.Thus, Area ofMNOP= 5 m x 5 m = 25 m2Hence,Area of the roads = Area (EFGH) + Area (JKIL) − Area (MNOP) = (350 + 225 ) m2− 25 m2= 550 m2Again, it is given that the cost of constructing the roads = Rs. 105 per m2Therefore,Cost of constructing 550 m2area of the roads = Rs. (105 × 550) = Rs. 57750. |
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| 63119. |
(a) 5p+2=DWhat number is 3 more than the product of two and four?C?hot is the numb |
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Answer» 2*4 = 8Then, 8+3 is 11. |
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| 63120. |
-14710. The product of twoers is(-27) . If one of the number is 5 , find the other numb37 |
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Answer» the verification is to know whether your answer is right |
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| 63121. |
A rectangular lstrrp ett ottawn 70 m by 50 m has two cross roads each 6 m wide running in the middle of it, orethe length and other parallel to the breadth. Find the cost of levelling them at ?5.50 per squaremetre. |
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| 63122. |
ctos |
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| 63123. |
оо28. A class room is 7m long, 6.5m wide and 4m high. It has one door 3m X 1.4m and threewindows measuring 2m x 1m.the rate of Rs 3.50 per m2.The interior walls are to be painted. Find the cost of painting at |
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Answer» Area of four walls=(2h(l+b))=(2x4(7+6.5))=(8x13. 5)=108m squarearea of one window =3mx1. 4m=4.2m squarearea of three window =3(2mx1m)=6m squaretotal area of windows =(4.2+6)m square=10.2m squareArea for whitewashing =(108-10.2)m square=97.8m square cost of 1 m square white washing = 3.50rupeecost of 97.8m square white washing=3.50x97.8=342.3 rupees |
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| 63124. |
A rectangular park is 90 m long and 70 m broad. Apath 6 m wide is to be built around the inside of thepark. Find the area of the path. Also, find the areaof the remaining park in hectares. |
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| 63125. |
3. A rectangular park is 90 m long and 70 m broad. Apath 6 m wide is to be built around the inside of thepark. Find the area of the path. Also, find the areaof the remaining park in hectares |
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Answer» Length of the rectangular park = 90 mBreadth of the rectangular park = 70 mWidth of the path = 6 mBreadth of the path = 70 - ( 6 x 2 )= 70 - 12= 58 mLength of the path = 90 - ( 6 x 2 )= 90 - 12= 78 mArea of the path = l x b= 78 m x 58 m= 4524 m²Area of the park including thepath = l x b= 90 m x 70 m= 6300 m²Area of the park excluding thepath = Area of the park including the path - Area of the path= 6300 m²- 4524 m²= 1776 m²1 hectare = 10,000 mArea of remaining park inhectares = 1776/10000= 0.1776 m²Area of the path is 4524 m². And the area of the remaining park in hectares is 0.1776 m². Like my answer if you find it useful! |
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| 63126. |
3. A rectangular park is 90 m long and 70 m broad. Apath 6 m wide is to be built around the inside of thepark. Find the area of the path. Also, find the areaof the remaining park in hectares. |
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Answer» Area of path=2(90*6+70*6) m²=2(540+420) m²=2*960 m²=1920 m² Area of remaining park=90*70-(1920) m²=6300-1920 m²=4380 m²=0.438 hectares |
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| 63127. |
ERCISE 19.1Find the volume and surface area of the cuboid.(i) I = 25 cm, b = 10cm, h = 5 cmGNI-27 |
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| 63128. |
7. Through a rectangular field of length 90 m and breadth 60 m, two roads areconstructed which are parallel to the sides and cut each other at right angles throughthe centre of the fields. If the width of each road is 3 m, find0 the area covered by the roads.(ii) the cost of constructing the roads at the rate of乏110 per m2. |
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| 63129. |
the length andwidth of a carpet are 4 m and 3 m,1tively. Find its area. Also, find the cost of therespearpet at the rate of t 45 per square metre. |
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Answer» Area = Length × Breadth = 4 × 3 = 12 cm^2 Cost of carpet = ₹45 × 12 = ₹540 area is 12 and cost rate is $540 |
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| 63130. |
10. Expresseach of the following ratios in simplest form.ii. 85 paise to ? 5v. 35 days to 6 weeksi. 25 cm to 1 m15 m/s to 108 km/h |
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Answer» b) 15m/s/108km/h = (15*3600)/(108*1000)=15/(3*10)=1/2 |
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| 63131. |
rea of the lawn.length and breadth of a rectangular park are in the ratio 5:2. A 2.5-m-wide pathing all around the outside of the park has an area of 305 m2. Find the dimensions of |
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| 63132. |
Paonator i inxandobtained í. Find d thi natĂmal numb |
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Answer» thanks |
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| 63133. |
The number decreased by 7 is 26, find thenumbernumb |
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Answer» Suppose x-7= 25x= 25+7= 32that number will be 32 |
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| 63134. |
6. Two cross roads, each of width 10 m, cut at right angles through the centre of arectangular park of length 700 m and breadth 300 m and parallel to its sides. Find thearea of the roads. Also find the area of the park excluding cross roads. Give thanswer in hectares. |
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Answer» hit like if you find it useful |
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| 63135. |
Two cross roads each two metre wide, runs at right anglesthrough the centre of a rectangular park 72 m by 48 m, suchthat each is parallel to one of the sides of the rectangle. Findthe area of the remaining portion of the park. |
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Answer» Area of first crossroad which is rectangular = l×b Hence area = 72m × 2m= 144 m² Area of 2nd crossroad whuch is rectangular = 48m× 2m= 96 m² Area of middle part which is square =2m×2m= 4m^2 Area of crossroads= (144+96) m² - 4 m² =240 m² - 4 m²= 236 m² |
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| 63136. |
cross roads, each of width 10 m,cot at right anges through the cente of spark of length 700m and breadth 300 m and parilile toits sides Find theeca of the roads. Also find the area of the park excadng ctos rads Cive tearea oanswer in hectares. |
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| 63137. |
TAND AREAss roads, each of width 10 m, cut at right angles through the centre of angur park of length 700 m and breadth 300 m and paralle toits sides. Findtheof the roads. Also find the area of the park excluding crossroads. Give theTvo croFind theanswer in hectares. |
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Answer» find the following x 3/4 |
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| 63138. |
A rectangular park 54 m x 40 m has toroads in the centre running paralle2 m wideto the breadth. Find:Area of roads.(b) Area of the path excluding roads.Cost of laying grass at 5 per square metre |
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| 63139. |
path 1 m wide is built inside a square park of side 30 m along its sides. Find the areof the path. Calculate the cost of constructing the path at the rate of t 70 per m2.mnntinn tahle H 25 cm of the table cover i |
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Answer» side of square park (A) = 30m width of path = 1m side of inner park (a) = 30-1-1 = 28m cost of construction = 70 per m^2 Area of path = Area of square park - area of park inside the path = A^2 - a^2 = 30^2 - 28^2 = 900 - 784 = 116m^2 Cost of construction of path = 70×116 =8120 Thanks |
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| 63140. |
The length of a grassy field is 80 m and breadthis of 60 m. There are two cross roads, each ofwidth 4 metre run at right angles through thecentre of rectangular park, parallel to its sides.Find the area of the road of the path |
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| 63141. |
23)Arectangularlawn,70mby50mhastworoadseach 5 m wide running throughits middle, one parallel to its length and the other parallel to its breadth. Find thecost of constructing the roads at 110 per m2. |
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| 63142. |
the parkrectangular lawn 70 m by 50 m has two roadmiddle, one parallel to its length and the other parullel to itsconstructing the roads at ? 120 per m2t, As, each 5 m wide. rumning through itsbreadth. Find the cost of |
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| 63143. |
I- of a number is less than the original number by 10, the original numb(a) 30(b) 60(c) 36(d) 45 |
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| 63144. |
Find the perimeter of a rectangular garden whose length is 4 times its width and which has area equal to20736 sq.5. |
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Answer» Let the width be x ,then length is 4x.We know,Area of rectangle=l*bTherefore,area=4x*x20736=4x^2144=2xx=72.Therefore ,perimeter=2(l+b)=2(4*72+72)=2(288+72)=2*360=720 |
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| 63145. |
(a) 16160(b) 4931 (c) 18265 ()7. Find the smallest square number which is8. Find the smallest 6 digit number which is a perfectnumber which is exactly divisible by 6, 9 and 15.9. Anu thinks of a number and subtractsfrom it. She multiplies the result by 6 and obtainsnumber and subtractsa number that is 3 more than 5 times the original numb10.Sum of 2 numbers is 40. If one of them is 10 morethan 5 times the original number find the original number.S is 40. If one of them is 10 more than the other. Find the numbers.11. What number should be subtracted from-12. By what number we divideto get213-613.Divide the product ofandby their difference.214. Evaluate : - +- 415. Find 2 rational number between 0 and -1 by mean method.- 1 - 116.Find 10 rational numbers between --- & -2 3 |
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Answer» 7. sol smallest square number 3 8. sol 310. sol required number =40one number = 10other number =40-10= 30 |
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| 63146. |
27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n 27), then find the ratiooftheir9hterms. |
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Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)We can consider the 9th term as the m th term.Let’s consider the ratio these two AP’s m th terms as am: a’m→(2)Recall the nth term of AP formula, an= a + (n – 1)dHence equation (2) becomes,am: a’m= a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].now substitute the value of m as 9so the answer becomes120/95 |
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| 63147. |
27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their9terms. |
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Answer» Sn/sn =7n+1/4n+27 (given) Where Sn is sum of n terms of 1st AP and sn is sum of n terms of second AP. Sn =n/2(2A +(n-1)D) (A =first term, D =common difference) (Sum of AP formula) sn=n/2(2a+(n-1)d) (a=first term, d=common difference) (Sum of AP formula) Sn/sn=(2A+(n-1)D)/(2a+(n-1)d) 7n+1 /4n+27 = (2A+(n-1)D) / (2a+(n-1)d) As, we have to find ratio of their 9th term, let n =2(9)-1=17 So, equation=> 7(17)+1/4(17)+27=2A+(16D)/2a+16d 120/95 =A+8D /a+8d So, T9/t9 =120/95=24/19 Where T9 is 9th term of first AP and t9 is 9th term of second AP. Hit like if you find it useful! |
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| 63148. |
4.phe-measure-st-each-side-at |
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Answer» area of equilateral ∆ = √3/4× (side)² => √3/4× (side)² = √3=> side ² = 4=> side = √4 = 2 cm I did not understand. tell me what you didn't understand in this ? What We will do of √3 it gets cancelled out on both sides √3/4 * side² = √3 => side² = √3*4/√3 .... ( both √3 cancel each other in the numerator and denominator) => side² = 4 |
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| 63149. |
1. Convert the following units.a) 1cm2b)mm2m2 1 hectarec) 12,075 hectares- |
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Answer» 1. 100mm^22. 10000m^23. 1.2075×10^8m^2 |
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| 63150. |
Two cross roads, each of width 5 m, run at right angles through the centreof a rectangular park of length 70 m and breadth 45 m and parallel toitssides. Find the area of the roads. Also find the cost of constructing theroads at the rate of 105 per m2 |
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