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2 918
#Day_13_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
A shopkeeper sold four articles – P, Q, R, and S
after giving two successive discounts.
Article P: Marked up by 50% above its cost price
and sold after discounts of Z% and 10%
respectively. Total discount offered is Rs. 336
and selling price after discount of Z% is 20%
more than the cost price.
Article Q: Sold after discounts of (L – 5) % and
6.66% respectively and ratio of selling price after
first discount to that of cost price is 5:4.
Difference between marked price and cost price
is 40Z, and the total discount offered on the
article is 25% less than the cost price of article P.
Article R: Selling price after first discount is 75%
more than cost price and article sold after
discounts of Y% and (200/7) % respectively.
Marked price of the article is twice as that of the
cost price.
Article S: Marked price of article is Rs. 600 more
than its cost price and article sold after discounts
of K% and 1.2Y% respectively. Total discount
offered is Rs. 308 which is Rs. 8 more than that
of article R. Selling price after first discount is
52% more than that of its cost price.
sиαρσnє🌱
1) Z is how much % more or less than that of Y?
a) 75%
b) 60%
c) 40%
d) 80%
e) None of these
2) A = Profit % earned on article Q. Only
numerical value
B = profit % earned on article S. Only numerical
value
Find the value of 15% of (A x B)?
a) 73
b) 146
c) 96
d) 112
e) None of these
3) For article R, if the article is marked up by
7Y% and sold after a discount of (10K/3) %, then
find its new cost price. Selling price of the article
remains the same.
a) Rs. 300
b) Rs. 360
c) Rs. 400
d) Rs. 320
e) None of these
4) C = Difference between marked price of P and
S
D = Difference between marked price of Q and R
Find the difference between C and D.
a) 400
b) 200
c) 600
d) 1000
e) None of these
5) Cost price of article R and S together is how
much % more or less than the combined cost
price of P and Q together?
a) 20%
b) 25%
c) 30%
d) 40%
e) None of these
2 918
Detailed Solution:
Male population in P = 200
So, 25% of total male population of all given
cities together = 200
So, total male population in all given cities
together = 200 x 4 = 800
Males in city Q = 800 x 45% = 360
Difference between Males in R and S = 10% x
800 = 80………... (1)
Also, sum of males in R and S = 800 – 360 –
200 = 240………. (2)
On solving both equations, we get
Males in R = (240 + 80)/2 or (240 – 80)/2 = 160
or 80
Males in S = 80 or 160
Female population in R is 175% more than that
of males
So, females in R = 275% x 80 or 275% of 160 =
220 or 440
So, total population of R = 80 + 220 or 160 +
440 = 300 or 600
From chart of total population,
(2M + N) + (6N + 1) + (2M – 1) + 5M = 100
9M + 7N = 100
N = (100 – 9M)/7
If M = 1, N = 13
N = (100 – 9M)/7 =91/7
Then total population of city S = 5M% = 5%
Total population of city P = (2M + N) % = 2 x 1 +
13 = 15%
Not possible because the total population of city
S is highest.
There is no common factor in 9 and 7, so next
number which is divisible by 7 is = 91 – 9 x 7 =
28
If M = 8, N = 28/7 = 4
No other possible values of M and N
So, population of city R = (2 x 8 – 1) % = 15% of
total population of all cities together.
Total population of all cities together = 300/15 x
100 or 600/15 x 100 = 2000 or 4000
Total population of city S = 40% x 2000 or 40%
or 4000 = 800 or 1600
Total population of city S < 1000, so possible
value of total population of all cities together =
2000
City %of total Total Males Females
population population
P 20% 400 200 200
Q 25% 500 360 140
R 15% 300 80 220
S 40% 800 160 640
sиαρσnє🌱
1.
According to question,
Males of city R = 80
Females of city S = 640
Ratio of females of S and males of R = 640:80 =
8:1
Required change = (8 – 1)/1 = 7 times (B)
2.
For city Q,
Males those age is less than equal to 30 years =
60% x 360 = 216
Females those age is less than equal to 30
years = 140 x 40% = 56
Required sum = 216 + 56 = 272 (C)
3.
For city S,
Total people surveyed = 800
People likes at least one brand = 800 x 75% =
600
People likes exactly two brands = 2/3 x 600 =
400
People likes exactly one brand + all three
brands = 600 – 400 = 200
People likes exactly all three brands x 4 =
people like exactly one brand.
So, number of people likes all three brands = 1/5
x 200 = 40 (A)
4.
I. Females in city Q is least among the given
cities.
Females in city Q = 140 (Least among the given
cities.)
This statement is true
II. Central angle belongs to total population of
city Q is 90 degrees.
Central angle belongs to total population of city
Q = 25/100 x 360 = 90 degrees
This statement is true
III. Males and females in city P is same.
Males in city P = Females in city P = 200
This statement is true.
So, all given statements are true. (C)
5.
Value of A = 1/10 x [360 – 140] = 22
Value of B = 1/10 x [640 – 160] = 48
Value of [1/10 x (A + B)]123 = 7123
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
So, cyclicity of unit digit of 7 = 4
123/4 = remainder is 3
So, required unit digit = 7^3 = 3 (A)
2 918
2) For city Q, 40% of males have age above 30
years, while 60% of female have age above 30
years age, then find the population of city those
age is less than equal to 30 years?
a) 228
b) 248
c) 272
d) 292
e) None of these
3) For city S, survey conducted about liking of
three brands – A, B, and C. Two third of people
those likes at least one of the brands, likes
exactly two brands, while people those likes
exactly one brand is four times of those likes all
three brands. Find the number of people those
likes all three brands, if 25% of total number of
surveyed people don’t likes any of the brand?
a) 40
b) 80
c) 20
d) 60
e) Can’t be determined
sиαρσnє🌱
4) Which of the following statement (s) is/are
definitely true?
I. Females in city Q is least among the given
cities
II. Central angle belongs to total population of
city Q is 90 degrees.
III. Males and females in city P is same.
a) I only
b) II and III only
c) I, II, and III
d) I and III
e) None of these
5) A = 1/10 of difference between males and
females in city Q
B = 1/10 of difference between males and
females in city S
Find unit digit of [1/10 x (A + B)]123
a) 3
b) 1
c) 9
d) 7
e) None of these
2 918
+1#Day_12_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
The chart given in pic shows the % distribution of total population of four cities – P, Q, R, and S and %
distribution of the number of males in these cities.
Note: a) Difference between male population of R and S is 10% of total male population, while female
population in R is 175% more than that of male population.
b) Male population of city P is 200, whereas M and N are natural numbers. Total population of city S is
highest (< 1000) among the given cities.
sиαρσnє🌱
1) Females of city S is how many times more
than that of males of city R?
a) 8
b) 7
c) 5
d) 6
e) Can’t be determined
2 918
Detailed Solution:
Let total number of students = 1200
Total number of boys = 400
So, total number of Girls = 1200 – 400 = 800
Number of girls in school S = 25% x 800 = 200
So, number of girls in school Q = 25% x 60%of
800 = 15% x 800 = 120
From 2nd pie chart,
25% + 15% + d + f = 100%
So, (d + f) = 60%
Number of girls in school R is maximum, that
means girls in school R > 25%
Also, f is a multiple of 10.
If f = 30%, then d = 30% [not possible, because
number of girls in schoolR is more than other
schools.
If f = 40%, then d = 20%
Value of b is 37.5% more than that of a
So,
Total Number of students in school Q and R is
8:11
That means, number of students in school R >
40% x 800 (=320)
320/1200 = 26% approx.
That means total number of students in school R
= 33% or 44% of total students (because value of
b is multiple of 11).
If total number of students in school R = 44%
Then number of students in school Q = 8 x 4 =
32%
Number of students in schools (P + Q + R) = 44 +
32 + 25 = 101% (NOT possible)
If total number of students in school R = 33%
So, number of students in school Q = 24%
So, number of students in school S = 100- 25 –
24 – 33 = 18%(possible)
So, Value of a= 24, value of b = 33 and value of c
= 18
If f = 50, then number of students in school R
>33% of total students which is not possible.
So, we have only one possibility.
Value of f = 40, value of d = 20
sиαρσnє🌱
School Total no. of students Girls Boys
P 300 160 140
Q 288 120 168
R 396 320 76
S 216 200 16
1. For School P,
Total number of boys in the school = 140
Number of students has average age of 20 years
= 80% x 140 = 112
Rest 28 students have distinct integral age.
We need to maximize the age of elder one. So,
80% students should have minimum age (20
years).
Out of rest 28 boys, one of the youngest boys
must have = 20 years.
So, age of rest 27 students = 21, 22, 23, 24, 25,
26, 27…….to 47.
So, age of eldest boy = 47 years (C)
2. Required difference = 168 – 76 = 92 (B)
3. Required % = 18/25 x 100 = 72% (D)
4. Required difference = 320 – 16 = 304 (A)
5. Total number of boys in school R = 76
Number of boys likes only one sports = 16 + 14 +
10 = 40
Number of boys likes (exactly two sports + all
three sports) = 76 – 40 = 36
Number of boys like exactly two sports, should be
multiple of = 9 + 10 + 8 = 27
We have only one possibility, number of boys
those likes exactly two sports = 27
So, number of boys likes all three sports = 36 –
27 = 9 (A)
2 918
2) Find difference between number of boys in school Q and that of R.
a) 76
b) 92
c) 46
d) 56
e) None of these
3) Total number students in school S are what % of total number students in school P?
a) 64%
b) 75%
c) 68%
d) 72%
e) None of these
sиαρσnє🌱
4) Number of girls in school R is how much more/less than the number of boys in school S?
a) 304
b) 294
c) 312
d) 324
e) None of these
5) For boys in school R, each boy likes at least one of the three games A, B, and C. Number of
boys those likes only A is 16, only B is 14, and only C is 10. Number of boys those likes only B&
C, only C& A, and only A& B are in ratio of 9:10:8 respectively. Find number of students those likes
all three games.
a) 9
b) 3
c) 6
d) 12
e) Can’t be determined
2 918
+1#Day_11_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
The pie chart given in pic shows the % distribution of total students in four different schools namely P, Q,
R and S and also% distribution of girls in these four different schools.
Note:
a) Total number of boys in all given schools together is 400 and value of b is 37.5% more than that of a.
b) Number of girls in school R is more than that in other schools and f is in exact multiple of 10.
c) Girls in school Q is 60% as that in S.
sиαρσnє🌱
1) For school P, 80% of boys have average age of 20 years and rest 20% are elder in the age as
compared to other boys (distinct integral age). Find the maximum possible age of eldest boy in
the school.
a) 32 years
b) 48 years
c) 47 years
d) 30 years
e) None of these
2 918
Detailed Solution:
1. For tank P,
Effective time taken by B to fill one tank = (106.5
– 1.5)/2.5 = 42 hours
So, time taken by Pipe C to empty the tank = 42 x
2 = 84 hours
Now,
2 x [(1/36 + 1/42 – 1/84)] x volume of tank = 540
10/252 x volume of tank = 270
So, volume of tank P = 6804 cm^3
Height of tank P = 21 cm
Now,
1/3 x 3 x K x K x 21 = 6804
So, value of K = 18 cm
Required value = 2 x 18 + 14 = 50 (B)
sиαρσnє🌱
2. For tank Q,
Time taken by pipe A alone to fill the tank = 16
hours
On every third hour, pipe C is opened alone. Tank
is filled in 10. 5 hours.
So,pipe C opened for 3 hours, and pipes A and B
opened for 7.5 hours.
Ratio of time taken by pipe B and C = 3:8
Now,
7.5 x 1/16 + 7.5 x 1/3a – 3 x 1/8a = 1
17/8a = 17/32
So, value of a = 4
Time taken by pipe C to empty the tank alone = 8
x 4 = 32 hours
So, rate of emptying by pipe C = (3 x 24 x 24 x
16)/32 = 864 cm^3/h (C)
sиαρσnє🌱
3. For tank R,
When A and C opened simultaneously, so tank is
never filled.
So, efficiency of C ≥ Efficiency of A
So, minimum possible efficiency of C = Efficiency
of A
So, time taken by C to empty the tank = 96 hours
So, time taken by B to fill the tank = 96/2 x 3 =
144 hours
Now,
3 x (1/96 + 1/144) x volume of tank = 480
So, volume of tank = 9216 cm^3
Height of the tank = 12 cm
Let breadth of the tank = 3a cm
So, length of the tank = 3a x 4/3 = 4a
Now,
12 x 3a x 4a = 9216
Value of a = 8
So, length of the tank = 8 x 4 = 32 cm (D)
4. For tank S,
Time taken by pipe B to fill the tank = 50 hours
So, time taken by pipe C to empty the tank =50 ×
6/5 = 60 hours
Efficiency of pipe C = (3 x 20 x 20 x 24)/60 = 480
units/hour
Since pipe A is filling and Pipe C is emptying and
efficiency of pipe C>A,
So, efficiency of pipe A = 480 – 192 = 288
units/hour
Required time = 60% x [(3 x 20 x 20 x 24)/288] =
60 hours (B)
sиαρσnє🌱
5. According to question,
5/72 x (3 x 20 x 20 x 24) = 2/3 x 3 x r3
So, r^3 = 1000
Radius of hemisphere tank = 10 cm (C)
