1. A shopkeeper purchased a product for Rs.$100$ and sells it
making a profit of $10$%. The customer resells it to the same shopkeeper
incurring a loss of $10$%. In these dealings the shopkeeper makes
1. no profit, no loss
2. Rs. $11$
3. Re. $1$
4. Rs. $20$
1. no profit, no loss
2. Rs. $11$
3. Re. $1$
4. Rs. $20$
Solution: CP for the customer: $100+10=110$
and $10$% of $110=11$ So customer sells shopkeeper at $99$. SO
Shopkeeper makes a profit of Rs $10+1=11$.
$2$nd option is correct.
2. A vessel is partially
filled with water. More water is added to it at a rate directly proportional to
time. Which of the following graphs depicts correctly the variation of total
volume $V$ of water with time $t$?
3. The triangle formed by the
lines $y=x$ $y=1-x$ and $x=0$ in a two dimensional plane is (x and y axes have the same scale)
1. isosceles and right-angled
2. isosceles but not right-angled
3. right-angled but not isosceles
4. neither isosceles nor right-angled
1. isosceles and right-angled
2. isosceles but not right-angled
3. right-angled but not isosceles
4. neither isosceles nor right-angled
Solution Slope of $y=x$ is $45$ degree. Slope of $y=1-x$ is $180-45$
degree. So In the triangle, two angles are $45$ and $45$. So $1$st option is
correct.
4. Statement A. The following
statement is true
Statement B. The preceding statement is false.
Choose the correct inference from the following:
Statement B. The preceding statement is false.
Choose the correct inference from the following:
1. Statements A and B are
always true
2. Statements A and B can be true if there is at least one statement between A and B
3. Statements A and B can be true if there are at least two statements between
A and B
4. Statements A and B can never be true, independently.
2. Statements A and B can be true if there is at least one statement between A and B
3. Statements A and B can be true if there are at least two statements between
A and B
4. Statements A and B can never be true, independently.
The number of squares in the
above figure is
1. $30$
1. $30$
2. $29$
3.$25$
4.$20$
6. A person walks downhill at
$10$ km/h, uphill at $6$ km/h and on the plane at $7.5$ km/h. If the person
takes $3$ hours to go from a place A to another place B, and $1$ hour on the
way back, the distance between A and B is
1. $15$ km.
2. $23.5$ km.
3. $16$ km.
4. Given data is insufficient to calculate the distance.
1. $15$ km.
2. $23.5$ km.
3. $16$ km.
4. Given data is insufficient to calculate the distance.
Solution: Data is not sufficient to solve it.
7. A bird leaves its nest and
flies away. Its distance x from the nest is plotted as a function of time t.
Which of the following plots cannot be right?
Sotution: $3$. is not possible . Since at one
time two value of $x$ is not possible.
8. A car is moving at 60
km/h. The instantaneous velocity of the upper most points of its wheels is
1. 60 km/h forward
2. 120 km/h forward
3. 60 km/h backward
4. 120 km/h backward
1. 60 km/h forward
2. 120 km/h forward
3. 60 km/h backward
4. 120 km/h backward
9. A living cell has a
protoplasm which is water based and demarcated by a lipid bilayer membrane. If
a cell is pierced up to $1/5$ th of its diameter with a very sharp needle, after taking the
needle out
1. no effect will be observed.
2. protoplasm will leak out from the hole made by the needle for a few minutes until the cell heals the wound.
3. protoplasm will keep on leaking out till the cell is dead.
4. the cell will burst like a balloon.
1. no effect will be observed.
2. protoplasm will leak out from the hole made by the needle for a few minutes until the cell heals the wound.
3. protoplasm will keep on leaking out till the cell is dead.
4. the cell will burst like a balloon.
Solution: No effect will be observed. (
Don't know how)
10. If D
+ I + M = 1501
C+ I + V + I + L = 157
L + I + V + I + D = 557
C +I + V+ I + C = 207
What is V + I + M = ?
1. Cannot be found
2. 1009
3. 1006
4. 509
C+ I + V + I + L = 157
L + I + V + I + D = 557
C +I + V+ I + C = 207
What is V + I + M = ?
1. Cannot be found
2. 1009
3. 1006
4. 509
Solution: Letters are written in Roman Numerals. So, Answer is
\(5+1+1000=1006\). So, third option is correct.
11. Density of a rice grain
is 1.5 g/cc and bulk density of rice heap is 0.80 g/cc. If a 1 litre container
is completely filled with rice, what will be the approximate volume of
pore space in the container?
1. 350 cc 2. 465 cc
3. 550 cc 4. 665 cc
1. 350 cc 2. 465 cc
3. 550 cc 4. 665 cc
Solution: Using allegation Formula:
Quantity of Cheaper/ Quantity of dearer= (high value-mean
value)/(mean value-low value)
volume of pour Space/Volume of rice=$\frac{1.5-0.8}{0.8-0}$=$\frac{7}{8}$
So volume of pour space=$1000/15\times 7$=$466.66$ approximately
$465$.
12. Four circles of unit
radius each are drawn such that each one touches two others and their centres
lie on the vertices of a square. The area of the region enclosed between the
circles is
1. $\pi-1$ 2. $\pi-2$
3. $3-\pi$ 4. $4-\pi$
1. $\pi-1$ 2. $\pi-2$
3. $3-\pi$ 4. $4-\pi$
13. A turtle starts swimming
from a point A located on the circumference of a circular pond. After swimming
for 4 meters in a straight line it hits point B on the circumference of the
pond. From there it changes direction and swims for 3 meters in a straight line
and arrives at point D diametrically opposite to point A. How far is point D
from A?
1. 3 m 2. 4 m
3. 7 m 4. 5 m
1. 3 m 2. 4 m
3. 7 m 4. 5 m
Solution: Thales theorem states
that any diameter of a circle Subtends a right angle to any point on
the circle.
So by applying Pythagoras's theorem , We get the required
distance=$5$
14. A film projector and
microscope give equal magnification. But a film projector is not used to see
living cells because
1. a living cell cannot be placed in a film projector.
2. the viewer’s eye is close to a microscope whereas it is far away from the projector’s screen.
3. a microscope produces a virtual image whereas a projector produces a real image.
4. a microscope has greater resolving power than a projector.
1. a living cell cannot be placed in a film projector.
2. the viewer’s eye is close to a microscope whereas it is far away from the projector’s screen.
3. a microscope produces a virtual image whereas a projector produces a real image.
4. a microscope has greater resolving power than a projector.
Solution: Resolving power is the ability of of an
optical instrument or type of film to separate or distinguish small or closely
adjacent images. SO $4$ option is correct.
15. In each of the following
groups of words is a hidden number, based on which you should arrange them in
descending order. Pick the correct answer:
E. Papers I Xeroxed
F. Wi-Fi veteran
G. Yourself ourselves
H. Breaks even
1. H, F, G, H 2. E, G, F, H
3. H, F, G, E 4. H, E, F, G
E. Papers I Xeroxed
F. Wi-Fi veteran
G. Yourself ourselves
H. Breaks even
1. H, F, G, H 2. E, G, F, H
3. H, F, G, E 4. H, E, F, G
Solution:
E. Papers I Xeroxed
F. Wi-Fi veteran
G. Yourself ourselves
H. Breaks even
So, correct order is H, E, F, G
E. Papers I Xeroxed
F. Wi-Fi veteran
G. Yourself ourselves
H. Breaks even
So, correct order is H, E, F, G
16. Five congruent rectangles
are drawn inside a big rectangle of perimeter 165 as shown. What is the
perimeter of one of the five rectangles?
Solution:
let length and breadth of one rectangle is $l$ and $b$ respectively. Then from this picture it's clear that
let length and breadth of one rectangle is $l$ and $b$ respectively. Then from this picture it's clear that
$3\times b=2\times l$ (see pic)
Also perimeter of big rectangle is $2\times
l+2\times(l+b)+3\times b=165$ Solving these two equation we get $l=3/2\times 15$
and $b-15$ So, perimeter =$2\times (l+b)=75$
17. At one instant, the hour
hand and the minute hand of a clock are one over the other in between the
markings for 5 and 6 on the dial. At this instant, the tip of the minute hand
1. is closer to the marking for 6
2. is equidistant from the markings for 5 and 6
3. is closer to marking for 5
4. is equidistant from the markings for 11 and 12
1. is closer to the marking for 6
2. is equidistant from the markings for 5 and 6
3. is closer to marking for 5
4. is equidistant from the markings for 11 and 12
Solution: $\theta=|30h-11/2m|$ So here hour hand is between $5$
and $6$ So $h=5$ and $\theta=0$ So we have $0=150-11/2 \times m$ That is, $m=25.4$
that means close to $5$. so correct option is $3$.
18. A cubical cardboard box
made of 1 cm thick card board has outer side of 29 cm. A tight-fitting cubical
box of the same thickness is placed inside it, then another one inside it and
so on. How many cubical boxes will be there in the entire set?
1. 29 2. 28 3. 15 4. 14
1. 29 2. 28 3. 15 4. 14
Solution: Side of $1$st Cube is 29 .
Inside it Cube will have a side =$27$. So sequence will be
$29,27,...,2$ So, Let number of terms in this sequence is $n$.
then $2=29+(n-1)\times (-2)$ and $n$ is a bit greater than $14$ But
here cube can be fractional So answer is $14$.
19. There are two buckets A
and B. Initially A has 2 litres of water and B is empty. At every hour 1 litre
of water is transferred from A to B followed by returning litre back to A from
B half an hour later. The earliest A will get empty is in:
1. 5 h 2. 4 h
3. 3 h 4. 2 h
1. 5 h 2. 4 h
3. 3 h 4. 2 h
Solution:
So Answer will be $3$ Hours.
20. Secondary colours are
made by a mixture of three primary colours, Red, Green and Blue, in different
proportions; each of the primary colours comes in 8 possible levels. Grey
corresponds to equal proportions of Red, Green and Blue. How many shades of
grey exist in this scheme?
1. $8^3$ 2. 8 3. $3^8$ 4. $8\times 3$
1. $8^3$ 2. 8 3. $3^8$ 4. $8\times 3$
Solution: every Shades have $8$ options to get selected to produce
required color. So Answer will be $8$
grey is the combination of equal proportions of red,green and blue answer will be 8 only.
ReplyDeletethanks
DeleteWhat do you mean by allegation formula?
ReplyDeleteQn.6 answer is 15 , if the distance of hill is x(same for up and down) and plane is y,
ReplyDeletewe get total time taken from A to B is 3 = x/6 + y/7.5 and from B to A is 1 = x/10 + y/7.5. solving these two equations we get x+y=15.
Qn 6 answer is 15
ReplyDeleteThank you Aprajita 😊
ReplyDeleteSolution exits for 6th question
ReplyDeleteAns is 15km
Soln
Consider uphill=a downhill =b onplane=c
Time=distance/speed
For 3hrs
Eqn
a/6 +b/10 +c/7.5 =3....i
For 1hr uphill=downhill down hill=uphill onplane remains same
Hence eqn becomes
b/6 +a/10 +c/7.5=1...ii
Solving i and ii
WE get
a-b=30....iii
Now i*3,ii*5
We get
2b + c=-15....iv
Now i*5 ii *3
We get
2a+ c=45..v
Add iii iv and v
3a+b+2c=60
a+b+c+(2a+c)=60
a+b+c=60-45
Total distance is 15kms
Thank you so much
ReplyDelete