1. In a vessel containiing, 60 litres of mixture of milk and water, water is only 30%. 30 litres of mixture is taken out and replaced with x litres of pure milk. As a result the percentage of water in the mixture becomes 20%. What is the value of x ?
A. 5
B. 10
C. 15
D. 8
E. 12
Sol. Amount of water initially = (30/100)*60 = 18 litres
Amount of milk initially = (60 - 18) = 42 litres
Now, 30 litres of mixture is taken out and replaced with x litres of milk.
Amount of water = 18 - (18/60)*30 = 9 litres
Amount of milk = 42 - (42/60)*30 + x = (21 + x) litres
Total quantity = 9 + 21 + x = (30 + x) litres.
Now its given that percentage of water is 20%. so,
(20/100)*(30+x) = 9
x = 15 litres.
2. Ramesh won a competition and got some prize money. He gave Rs. 2000 less than the half of prize money to Suresh and Rs. 1000 more than the two third of the remaining to his Mahesh. If both of them got the same amount, what is the prize money Ramesh got ?
A. 20000
B. 25000
C. 26000
D. 24000
E. None of these
Sol. Assume Ramesh got x rupees.
He gave x/2 - 2000 to Suresh.
And (2/3)*( x/2 + 2000 ) + 1000 to Mahesh.
x/2 - 2000 = x/3 + 7000/3
x = 26000.
Directions(3-7): Find the missing numbers in the givern series below:
3. 3, 4, 12, _, 576, 27648
A. 48
B. 64
C. 36
D. 72
E. None of these
Sol. 4*3 = 12
12*4 = 48
48*12= 576
576*48= 27648
4. 4, 12, 60, 420, _, 41580
A. 4620
B. 3780
C. 4200
D. 5040
E. None of these
Sol. 4*3 = 12
12*5 = 60
60*7 = 420
420*9 = 3780
3780*11= 41580
5. 686, 438, 314, 252, 221, _
A. 180
B. 187.5
C. 201
D. 205.5
E. 175.5
Sol. 686 – 248 = 438
438 – (248/2) = 438 – 124 = 314
314 – (124/2) = 314 – 62 = 252
252 – (62/2) = 252 – 31 = 221
6. 4, 3, 12, 6, 36, 9, 108, 12, _
A. 324
B. 252
C. 156
D. 182
E. 236
Sol. It is a double series.
*3 :– 4, 12, 36, 108
+3 :– 3, 6, 9, 12
7. 120, 170, 224, 280, 360, _
A. 392
B. 442
C. 458
D. 528
E. 540
Sol. 11^2 – 1 = 120
13^2 + 1 = 170
15^2 – 1 = 224
17^2 + 1 = 280
19^2 – 1 = 360
21^2 + 1 = 442
8. A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of while tiles is the same as the number of red tiles. A possible value of the number of titles along one edge of the floor is
A. 10
B. 12
C. 14
D. 16
E. None of these
Sol. Let the rectangle has m and n tiles along its length and breadth respectively.
The number of while titles
W = 2 m + 2 (n - 2) = 2(m + n - 2)
And the number of red tiles
R = mn - 2(m + n - 2)
Given
W = R ⇒ 4 (m + n - 2) = mn
⇒ mn - 4m - 4n = -8
⇒ (m - 4) (n - 4) = 8
As m and n are integers so (m - 4) and (n - 4) are both integers. The possibilities are (m - 4, n - 4) = (1 , 8) or (2, 4) giving, (m, n) as (5, 12) or (6, 8) so the edges can have 5, 12, 6 or 8 tiles. Answer is (b) only.
9. A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs 250 and Rs 300 per day respectively. In addition, a male operator gets Rs 15 per call he answers and female operator gets Rs 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 and maximum 12 number of the females?
A. 15
B. 14
C. 12
D. 10
E. None of these
Sol. Let us form both equations first:
40m + 50f = 1000
250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
850m + 8000 f = A
When M and F are the number of Males and Females and A is the amount paid by the service provider.
Then the possible values of F are 8, 9, 10, 11, 12
If F = 8, then, M = 15
If F = 9, 10, 11 then M will not be an integer while F = 12 then M will be 10.
By putting F = 8 and M = 15, A = 18800. When F = 12 and M = 10, then A = 18100.
Hence the number of males will be 10.
10. The cost price of goods with a bankrupt is Rs. 25500 and if the goods had realised in their full value, his creditiors would have received 85 paise in the rupee. But 2/5 of the goods were sold at 17% and the remainder at 22% below their cost price. How many paise in a rupee was received by the creditors?
A. 72 paise
B. 68 paise
C. 55 paise
D. 52 paise
E. None of these
Sol. Total debt=25500× 100/85=Rs.30000
Money received by selling the goods=25500(2/5×83/100+3/5×78/100)
=25500/500 (166+234)
=51×400=Rs.20400
Therefore, money received by the creditors for a rupee=Rs.(20400/30000)=Rs.0.68=68 paise
Hence, the creditor received 68 paise in a rupee.
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