331, 482, 551, 263, 383, 362, 284
Question: 331, 482, 551, 263, 383, 362, 284
[A].
263
383
331
551
[A].
[B].
[C].
[D].
Answer: Option B
Explanation:
In each number except 383, the product of first and third digits is the middle one.
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The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
Question: The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
[A].Rs. 1200 [B].
Rs. 2400 [C].
Rs. 4800 [D].
Cannot be determined Answer: Option B
Explanation:
Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 …. (i)
and x + 6y = 1600 …. (ii)
Divide equation (i) by 2, we get the below equation.
=> x + 2y = 800. — (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 (-)
x + 2y = 800
—————-
4y = 800
—————-Therefore, y = 200.
Now apply value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400
Solving (i) and (ii) we get x = 400, y = 200.
Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
Question:
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
[A].4 days [B].
5 days [C].
6 days [D].
7 days Answer: Option A
Explanation:
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1 and 26x + 48y = 1 . 10 2 Solving these two equations, we get : x = 1 and y = 1 . 100 200 (15 men + 20 boy)’s 1 day’s work = 15 + 20 = 1 . 100 200 4 15 men and 20 boys can do the work in 4 days.
The banker’s discount on a bill due 4 months hence at 15% is Rs. 420. The true discount is:
Question:
The banker’s discount on a bill due 4 months hence at 15% is Rs. 420. The true discount is:
[A].Rs. 400 [B].
Rs. 360 [C].
Rs. 480 [D].
Rs. 320 Answer: Option A
Explanation:
T.D. = B.D. x 100 100 + (R x T) = Rs. 420 x 100 100 + 15 x 1 3 = Rs. 420 x 100 105 = Rs. 400. A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will he take to go 5km in stationery water?
A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will he take to go 5km in stationery water?A. 40 Minutes
B. 1 Hour
C. 1 hr 15 Min
D. 1 Hr 30 Min
Explanation:
Speed upstream = 2km/hr.
Speed downstream = 6 km/hr
Speed in stationary water = ½ (6 + 2) km/hr = 4 km/hr
Time is taken to cover 5 km in stationary water
= (1/4 x 5) 1 hr 15 minWhich of the following statements is not correct?
Question: Which of the following statements is not correct?
[A].log10 10 = 1 [B].
log (2 + 3) = log (2 x 3) [C].
log10 1 = 0 [D].
log (1 + 2 + 3) = log 1 + log 2 + log 3 Answer: Option B
Explanation:
(a) Since logaa = 1, so log10 10 = 1.
(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3
log (2 + 3) log (2 x 3)
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.
So, (b) is incorrect.
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
Question:
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
[A].35 [B].
36 2 3 [C].
37 1 2 [D].
40 Answer: Option D
Explanation:
Let distance = x km and usual rate = y kmph.
Then, x – x = 40 2y(y + 3) = 9x ….(i) y y + 3 60 And, x – x = 40 y(y – 2) = 3x ….(ii) y -2 y 60 On dividing (i) by (ii), we get: x = 40.