A. 12 π cm3
B. 6 π cm3
C. 8 π cm3
D. 10 π cm3
A. 66
B. 48
C. 24
D. 8
Let the number of 1 rupee, 2 rupee and 5 rupee coins be 3x, 5x and 4x. (1*3x)+(2*5x)+(5*4x)=264 3x+10x+20x=264 33x=264 x=8 No. of 1 rupee coins
= 3x = 24
A. 5 5⁄11
B. 4 5⁄11
C. 6 4⁄11
D. 6 5⁄11
A can complete the work in 12 days working 8 hours a day
=> Number of hours A can complete the work = 12×8 = 96 hours
=> Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day
=> Number of hours B can complete the work = 8×10 = 80 hours
=> Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480
=> A and B can complete the work in 480/11 hours
A and B works 8 hours a day
Hence total days to complete the work with A and B working together
= (480/11)/ (8) = 60/11 days = 5 5⁄11 days
A. Rs 17423
B. Rs 18973
C. Rs 19448
D. Rs 19880
Explanation:
P = Rs 16000, R = (10/2)%
Per half-year , t= 4 half-years
Therefore Amount = Rs [16000 × (1+5/100)4]
= Rs (16000 × 21/20 × 21/20 × 21/20 × 21/20)
= Rs 1448.10 ~ Rs 19448
A. 120
B. 80
C. 192
D. 48
If A invests amount C1 for T1 time and his of profit is P1, and B invests amount C2 for T2 time and his of profit is P2, then, C1 * T1 / C2 * T2 = P1/P2
If P is the Nasir’s of profit, then Changaz gets (240 – P)
Therefore, 6000 * 12 / 3000 * 6 = (240 – P) / P = 72/ 18 = 4
4P = (240 – P)
5P = 240
P = 48
A. 200 sq cm
B. 72 sq cm
C. 162 sq cm
D. Cannot be determined
Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.
A. 35%
B. 38%
C. 32%
D. 29%
E. None of these
Explanation:
Let the CP of an article be Rs. 100
Given that successive discounts are 20% and 15%.
SP = 85% of 80% of 100 = (85/100)(80/100)(100)
=> SP = Rs. 68
Clearly, single discount is 32%.