An engineering student has to secure 36% marks to pass. He gets 130 marks and fails by 14 marks. The maximum No. of marks obtained by him is_________?
A. 300
B. 400
C. 350
D. 500
A. 300
B. 400
C. 350
D. 500
A. 216
B. 217.50
C. 236.50
D. 245
Explanation:
Given expression = (45/100 * 750) – (25/100 * 480) = (337.50 – 120) = 217.50
A. 600
B. 620
C. 500
D. 520
E. None of these
Explanation:
Let the number of children in the school be x. Since each child gets 2 bananas, total number of bananas = 2x.
2x/(x – 360) = 2 + 2(extra)
=> 2x – 720 = x => x = 720.
A. 60
B. 64
C. 72
D. 80
Explanation:
Given M + P + C = 80 * 3 = 240 — (1)
M + P = 90 * 2 = 180 — (2)
P + C = 70 * 2 = 140 — (3)
Where M, P and C are marks obtained by the student in Mathematics, Physics and Chemistry.
P = (2) + (3) – (1) = 180 + 140 – 240 = 80
A. 28
B. 32
C. 40
D. 64
Explanation:
Let the numbers be 2x and 3x.
Then, their L.C.M = 6x. So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
A. Rs. 15000
B. Rs. 16200
C. Rs. 14700
D. Rs. 15900
E. None of these
Explanation:
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000