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M4maths Previous Puzzles

30 Sep 2022 Area & Volume

When a cylindrical tank is 3/7 full, an inlet pipe with a capacity of 4928 liters per hour fills it completely in 4 hours. If the height of a tank is two times its diameter, then the radius of a tank is: 

OPtion

1) 2.8 m
2) 1.9 m
3) 2.1 m
4) 2.4 m
5) 1.4 m
6) 2.0 m
7) 1.8 m
8) 3.5 m
9) 1.2 m
10) None of these
Solution
4/7 volume of a tank = 4928x4 = 19712 liters
.'. Total capacity of tank = (7/4)x19712 = 34496 liters
⇒ Volume of a cylindrical tank = πr²h = 34496 litres = 34496000 cm³
Given, h=2xDiameter=2x(2r)=4r
.'. πr²(4r) = 34496000
(22/7)x4xr³ = 34496000
r³ = 2744000
r = 140 cm = 1.4 m
Correct Option 5)
Correct Option: 5 Best Solution (0)
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29 Sep 2022 Time & Work

20 women work for 4 days on a certain project and complete one-tenth of it. 15 men work for the next 'X' days and complete one-fifth of it, If after (4+X) days, 20 women and 15 men work together and complete the remaining work in 12 days, then how many days will 15 men take to complete the whole work ?

OPtion

1) 24
2) 30
3) 45
4) 60
5) 36
6) 48
7) 18
8) 27
9) 42
10) None of these
Solution
20 Women's 1 day work=1/(4*10)=1/40
15 men's 1 days work=1/(X*5)=1/(5X)
After total (4+X) days, work comleted=1/10 + 1/5=3/10 and remaining work=7/10
As 20 women and 15 men complete the remaining work in 12 days, so their 1 day work=(1/12)*(7/10)=7/120
Comparing 1 day's work of 20 women and 15 men=1/40 + 1/(5X)=7/120
(X+8)/(40X)=7/120
Solving, we get X=6
.'. 1 day's work of 15 men=1/(5*6)=1/30
Hence, 15 men can complete the whole work in 30 days.
Correct Option 2)
Correct Option: 2 Best Solution (1)
G.S.Kantharao   6 Months AGO
20w - 4d -1/10 =>in 1d=1/40
15m- Xd - 1/5 => in 1d=1/5x
Remaining work =1-(1/10+1/5)=1 -3/10= 7/10
20w+15w - 12d - 7/10
12(1/40+1/5x)=7/10
12(x+8)/40x =7/10
12x+96=28x=>16x=96=>x=6
So 15m - 6d -1/5=>1d=1/30
Full work can be completed in =30d
My option is 2👈
Like? Yes (1) | No   

28 Sep 2022 Time Distance & Speed

A boy started for a school which is 12 km from his home by bicycle at a speed of 8 km/hr but after 15 minutes he remembers that he forgot one of the book at home, so he increases his speed and returns back to home and also reaches to school in time. The percentage increase in his speed is

OPtion

1) 33.33%
2) 20%
3) 15%
4) 30%
5) 12.5%
6) 37.5%
7) 40%
8) 25%
9) 66.55%
10) None of these
Solution
Usual time to reach school=12/8=1.5 hrs
Distance covered in 15 min. at a speed of 8 km/hr=(15/60)*8=2 km
So, he comes back 2 km to home and again return to school at a speed of say 'x' km/hr.
Then, usual time of 1.5 hrs = 15 min. + Time taken for the journey of (2+12) km with increased speed.
1.5 = (1/4) + (14/x)
x=11.2
Therefore % increase in the speed = [(11.2 - 8)/8]*100 = 40%
Correct Option 7)
Correct Option: 7 Best Solution (4)
G.S.Kantharao   6 Months AGO
Regular time to be taken =12/8
After 15 m he traveled 2km
So real distance to travel is =12+2=14km
Real time taken
=14/s=(12/8 - 1/4)=14/s=10/8
S=11.2kmph
%of increase=(11.2-8)100/8=40%
My option is 7 👈
Like? Yes (2) | No   
AAKASH RAWAT   6 Months AGO
Boy to school distance 12 km
Speed of bicycle 8 km/ hr
Time taken to go to school = 12/8 = 3/2hr
After 15 minutes ,
d = (15/60) ×8 = 2km
Now he will travel 14km to reach to school
Time left to reach school = (3/2) - 15/60 = 5/4 hr
Now speed will be 14/(5/4) = 56/5 km/ hr
percentage change in speed = (56/5)/8 ×100 = 140%
Increase in speed( 140 - 100)% intial speed 100%
= 40%
Like? Yes (2) | No   
RAMADURAI NATESAN   6 Months AGO
Initial time to reach school = 90 minutes.
Distance travelled in 75 m =14 km.Speed = 60*14/75 = 11.20km
% increase = 40
Like? Yes | No   
Dr.Dipin Singh   6 Months AGO
Boy will take 1.5 hrs to cover 12 kms at speed of 8 km/hr.
In 15 minutes, he covered 2 kms.
Now to reach school in time, he should cover 14 kms in 1.25 hrs or 5/4 hrs.
So new speed =14*4/5=56/5 kmph which is 7/5 or 1.4 times of earlier speed.
So increase in speed=40%
Like? Yes | No   

27 Sep 2022 Seating Arrangement

Six friends, A, B, C, D, E and F are sitting in two rows, three in each. D is not at the end of any row. E is a neighbour of C. C and F are sitting diagonally opposite to each other. A is second to the left of F. Who among the following three friends are in the same row?

OPtion

1) CDB
2) EAF
3) BDF
4) ACF
5) BCF
6) EDF
7) EBA
8) BEC
9) CDF
10) None of these
Solution
According to the information given, the sitting arrange of A,B,C,D,E and F row-wise is:
C - E - B
A - D - F
So, from the given options Only B, E and C will be in one row.
Correct option 8)
Correct Option: 8 Best Solution (2)
G.S.Kantharao   6 Months AGO
row-1 C E B
row-2 A D F
So my option is 8👈
Like? Yes (1) | No   
RAMADURAI NATESAN   6 Months AGO
Following seating Arrangement Satisfies the given conditions
B E C
F D A
Like? Yes | No   

26 Sep 2022 Sequence & Series

Select the correct option which can replace question mark (?) in the given series
ZC29, VE27, RI27, NO29, ?

OPtion

1) IW32
2) JX34
3) HW33
4) JX32
5) KV32
6) JW34
7) IX33
8) JW33
9) JV32
10) None of these
Solution
When letters from A to Z are numbered from 1 to 26 respectively.
The sequence of position of first letter Z(26), V(22), R(18), N(14),..decrease by 4, so next letter at position 10=J
The second letter position C(3), E(5), I(9), O(15),... Increasing difference of 2, 4, 6,.., so next difference will be 8 and position 23=W
Two-digit number after letters is obtained by the sum of positions of first two letters i.e. ZC(26+3=29), VE(22+5=27), RJ(18+9=27), NO(14+15=29), so next number will be JW(10+23=33)
Hence correct cluster is JW33
Option 8)
Correct Option: 8 Best Solution (0)

23 Sep 2022 Time Distance & Speed

A bus can travel 66.66% faster than a car. Both start from point X at the same time and reach point Y which is 114 kms away from X at the same time. On the way, however, the bus lost about 48 minutes while stopping at the stations. The speed of the bus (in km/hr) is:

OPtion

1) 100.5
2) 94.5
3) 59
4) 57
5) 76
6) 95
7) 75
8) 92
9) 87
10) None of these
Solution
Let speed of the car=x km/hr, then speed of the bus=(166.66/100)*x=1.6666x=5x/3
Time taken by bus while travelling + Time of stoppages = Time taken by car
114/(5x/3) + 48/60 = 114/x
(342 + 4x) / 5x = 114/x
4x = 228
x = 57
.'. Speed of the bus = 5x/3 = 5*57/3 = 95 km/hr
Correct Option 6)
Correct Option: 6 Best Solution (0)

22 Sep 2022 Time & Work

Chirag and Bhavesh working together on certain job takes 7½ days to complete it. Chirag alone takes 8 days less than Bhavesh alone takes. In how many days Akash alone can complete the same work, if he is 60% more efficient than Chirag?

OPtion

1) 15
2) 25
3) 12½
4) 18⅘
5) 22½
6) 18⅓
7) 15⅓
8) 7½
9) 6⅖
10) None of these
Solution
Let Bhavesh alone takes 'x' days, then Chirag takes (x-8) days to finish the work.
1 day's work of Chirag and Bhavesh = 1/(x-8) + 1/x = 2/15
⇒ 2x² - 46x + 120 = 0
(2x-6) (x-20) = 0
x=20
.'. Bhavesh takes 20 days and Chirag takes 12 days.
Hence days required by Akash = (100/160)*12 = 7.5
Correct Option 8)
Correct Option: 8 Best Solution (2)
Dr.Dipin Singh   6 Months AGO
From given conditions,
Chirag takes 12 days and Bhavesh 20 days to complete the work independently.
Akash takes 12/1.6 = 7.5 days to complete the work alone.
Like? Yes (1) | No   
G.S.Kantharao   6 Months AGO
Let Chirag and Bhavesh can do in x,x+8 days
1/x +1/(x+8)=2/15
2x+8/x^2+8x=2/15
x^2-7x-60=0=>x=12
Let Aakash=K
Now 12/K = 160/100=>K=7.5 days
My option is 8 👈
Like? Yes | No   

21 Sep 2022 Profit & Loss

In the Meena Bazar ladies purchases with a fair bargaining, so the shopkeepers markup the prices too much. One of the shopkeeper marked up an article at Rs. X expected huge profit if it is sold on the marked price. But a lady purchased it at X/2 with his fine bargaining skills, so the expected profit of the shopkeeper diminished by 66.66%. What is the percentage discount fetched by the customer through bargaining?

OPtion

1) 33.33 %
2) 44.44%
3) 66.66%
4) 22.22%
5) 16,66%
6) 50%
7) 200%
8) 100%
9) 150%
10) None of these
Solution
Let CP=C, Given MP=X .'. Profit without any discount=(X-C)
With discount, when SP=X/2, then Profit=X/2 - C
Given, profit after discount is 66.66 less or is 33.33 of the original profit.
⇒ (33.33/100)(X-C) = (X/2 - C)
(1/3)*(X-C) = (X-2C)/2
X=4C
⇒ MP is 4 times the CP.
When CP=100, MP=400 and after bargain SP=200
.'. % Discount fetched by customer = [(400-200)/400]*100 = 50%
Correct Option 6)
Correct Option: 6 Best Solution (0)

20 Sep 2022 Sequence & Series

For an Arithmetic Progression (AP), the sum of first four terms and first eight terms is 288. If the sum of first twelve terms is 480, then sum of first 15 terms of this AP is:

OPtion

1) 537
2) 832
3) 644
4) 595
5) 735
6) 559
7) 715
8) 635
9) 670
10) None of these
Solution
For an AP, if first term=a and difference=d, then Sum of first 'n' terms of an AP=Sn=(n/2)[2a + (n-1)d]
.'. S4 = (4/2)(2*a + 3d) and S8 = (8/2)(2a + 7d)
S4+S8 = (4/2)(2*a + 3d) + (8/2)(2a + 7d) = 288
⇒ 6a + 17d = 144 ---(i)
Similarly, S12 = (12/2)(2a + 11d) = 480
⇒ 2a + 11d = 80 ---(i)
Solving Eqn, (i) & (ii), we get a=7, d=6
Now S15=(15/2)(2*7 + 14*6)=735
Correct Option 5)
Correct Option: 5 Best Solution (3)
G.S.Kantharao   6 Months AGO
S4+S8=4(2a+3d)/2 + 8(2a+7d)/2=288
4a+6d+8a+28d=288=>
12a+34d=288
S12=12(2a+11d)/2=12a+66d=480
(66-34)d=480-288=192
32d=192=>d=6,a=7
S15=15(2*7+14*6)/2
=15*98/2=15*49=735
My option is 5 👈
Like? Yes (2) | No   
AAKASH RAWAT   6 Months AGO
S4+ S8 = 288••••(1)
on solving
6a +17d = 144
Given,
S15 = 480
on solving
2a +11d= 80•••••eqn 2
on solving eqn 1 and eqn 2
a= 7,d=6To find S15
so ,
15/2(14+ 14× 6)
735
Like? Yes (2) | No   
RAMADURAI NATESAN   6 Months AGO
S = n/2(a+(n - 1 ) d). S4 = 2(2a + 3d) and S8 = 4(2a + 7d ).
S4 + S8 =12a + 34d = 288. S12 = 6( 2a+11d) = 480.
Solving, a = 7 & d = 6.
S15 = 7.5 (98) = 735
Like? Yes (1) | No   

19 Sep 2022 Area & Volume

Length, breadth and height of a room are in the ratio 6:5:3. If a new room is constructed where length is increased by 40%, breadth by 20% but the area of all four wall is kept same as original room, then change in volume of a new room compared to original is

OPtion

1) 28.80%
2) 21.66%
3) 14.40%
4) 33.66%
5) 14.33%
6) 21.40%
7) 15.20%
8) 30.66%
9) 28.33%
10) None of these
Solution
For the original room, let Length=6, Breadth=5 and Height=3, then its area of 4 walls=2(6*3 + 5*3)=66 unit² and Volume=6*5*3=90 unit³
For a new room, Length=1.4*6=8.4, Breadth=1.2*5=6. If height is 'x' then area of 4 walls=2(8.4x + 6x)=28.8x unit²
As the area of walls of both rooms is same .'. 28.8x=66 ⇒ x=165/72
So, the volume of new room=8.4*6*(165/72)=115.5 unit³
Hence percentage increase in volume=[(115.5-90)/90]*100=28.33%
Correct Option 9)
Correct Option: 9 Best Solution (0)
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