Maths Questions for DSSSB Exam : 16th May 2018 (Solutions)_00.1
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Maths Questions for DSSSB Exam : 16th May 2018 (Solutions)

Maths Questions for DSSSB Exam : 16th May 2018 (Solutions)_40.1

Q1. Weight of a sumo is jointly varies as his height and his age. When height is 1.2 m and age is 20 years, his weight is 48 kg. Find the weight of the sumo when his height is 1.5 metre and age is 30 years: 
(a) 60 kg 
(b) 72 kg 
(c) 90 kg 
(d) 58 kg 
Q2. A couple got married 9 years ago when the age of wife was 20% less than her husband. 6 years from now the age of wife will be only 12.5% less than her husband. Now they have six children including single, twins and triplets and the ratio of their ages is 2 : 3 : 4 respectively. What can be the maximum possible value for the present age of this family? 
(a) 110 years 
(b) 103 years 
(c) 105 years 
(d) 83 years 
Q3. In Maa Yatri Temple every devotee offers fruits to the orphans. Thus every orphan receives bananas, oranges and grapes in the ratio of 3 : 2 : 7 in terms of dozen. But the weight of a grape is 24 gm and weight of a banana and an orange are in the ratio of 4 : 5, while the weight of an orange is 150 gm. Find the ratio of all the three fruits in terms of weight, that an orphan gets: 
(a) 30 : 25 : 14
(b) 180 : 150 : 82
(c) 75 : 42 : 90
(d) 71: 63 : 67
Q4.There are five boxes in a cargo hold. The weight of the first box is 200 kg and the weight of the second box is 20% higher than the weight of the third box, whose weight is 25% higher than the first box’s weight. The fourth box at 350 kg is 30% lighter than the fifth box. Find the difference in the average weight of the four heaviest boxes and the four lightest boxes.
(a) 51.5 kg
(b) 75 kg
(c) 37.5 kg
(d) 112.5 kg
Q5.A machine depreciates in value each year at the rate of 10% of its previous value. However, every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of the fourth year, the value of the machine stands at Rs 1,46,205, then find the value of machine at the start of the first year.
(a) Rs 1,90,000
(b) Rs 2,00,000
(c) Rs, 1,95,000
(d) Rs 2,10,000
Q6. An amount of money is to be divided among P, Q and R in the ratio of 3 : 5 : 7 respectively. If the amount received by R is Rs. 4,000 more than the amount received by Q, what will be the total amount received by P and Q together? 
(a) Rs. 8,000
(b) Rs. 12,000
(c) Rs. 16,000
(d) Cannot be determined 
Q7. A 180-metre long train crosses another 270-metre long train running in the opposite direction in 10.8 seconds. If the speed of the first train is 60 kmph, what is the speed of the second train in kmph? 
(a) 80
(b) 90
(c) 150
(d) Cannot be determined 
Q8. The ratio of the ages of a father and a son at present is 5 : 2. Four years hence the ratio of the ages of the son and his mother will be 1 : 2. What is the ratio of the present ages of the father and the mother?
(a) 3 : 4
(b) 5 : 4
(c) 4 : 3
(d) Cannot be determined 
Q9. The length of a train and that of a platform are equal. If with a speed of 90 km/hr the train crosses the platform in one minute. Then the length of the train (in m) is: 
(a) 500
(b) 600
(c) 750
(d) 900 
Q10. A train passes a 50 m long platform in 14 seconds and a man standing on the platform in 10 seconds. The speed of the train is: 
(a) 24 km/hr
(b) 36 km/hr
(c) 40 km/hr
(d) 45 km/hr
Solutions:
S1. Ans.(c)
Sol. W∝HA ≡w=kHA
        48= k×1.2×20
K = 2 
So, required weight = k×1.5×30=90 kg. 
S2. Ans.(b)
Sol. Let husband present age = x 
Wife present age = y 
A/q, 
(x-9)/(y-9) = 100/80
& (x+6)/(y+6) = 100/87.5
On solving both equation, we get 
X = 34 years 
Y = 29 years 
Now, maximum age of any child must be less than 9 years. 
Hence their ages can be (2 : 3 : 4)  4, 6, 8 years. 
So, maximum possible value for present age 
of family =  34+29+ (1×4+2×6+3×8)
= 103 years. 
S3. Ans.(a)
Sol. Weight of banana : orange = 4 : 5 
 Orange = 150 gm
Banana = 120 gm
Grape = 24 gm 
Orphan receives banana, orange & grapes in ratio = 3 : 2 : 7 
So required ratio of fruits by weight  
 = 3 × 120 : 2 × 150 : 7 × 24
= 30 : 25 : 14   
S4. Ans.(b)
Sol.
Weight of first box = 200 kg
Weight of second box
=200×125/100×120/100
=300 kg 
Weight of third box 
=200×125/100
=250 kg
Weight of fourth box = 350 kg
Weight of fifth box 
=350×100/70
=500 kg
Required difference 
=1/4×(500+350+300+250)-1/4×(200+250+300+350)
=1/4×1400-1/4×1100 
=350-275
=75 kg
S5. Ans.(b)
Sol.
Let value of machine at the start of first year was Rs. x
ATQ, 
x×90/100×95/100×90/100×95/100=1,46,205 
⇒x=(1,46,205×10,00,000)/(81×95×95)
⇒x=Rs.2,00,000
S6. Ans.(c)
Sol.
Let amounts of P, Q and R be 3x, 5x and 7x respectively.
∴ 7x – 5x = 4000
⇒ x = 2000
∴ Required answer = 8 × 2000
= 16,000
S7. Ans.(b)
Sol.
Let speed of second train is x km/h
∴ 180 + 270 = (60 + x) × 5/18×10.8
⇒ x = 90 kmph
S8. Ans.(d)
Sol.
Let present age of father= 5x
Present age of son = 2x
Let mother’s present age = y years
∴(2x+4)/(y+4)=1/2
⇒ y= (4x + 4)
∴ Required ratio =5x/(4x+4)
Here, we don’t know value of x
∴ can’t be determined
Maths Questions for DSSSB Exam : 16th May 2018 (Solutions)_50.1
S10. Ans.(d)
Sol. Distance travelled in 14 sec. = 50 + l 
Distance travelled in 10 sec = l 
So, speed of train 
=50/(14-10) m/sec 
=50/4×18/5km/hr = 45 km/hr
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