Question – 1
1.
| Column-A | Column-B |
|---|---|
| (a^2-ab+b^2)/(a^2+ab+b^2) when a = -b | (a^2-ab+b^2)/(a^2+ab+b^2) when a = b |
(Note: [a^2=a*a])
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:A
- Answer Explanation:When a =-b(a^2-ab+b^2)/(a^2+ab+b^2)=[(-b)^2-(-b)b+b^2]/[(-b)^2+(-b)b+b^2]=(b^2+b^2+b^2)/(b^2-b^2+b^2)= 3b^2/b^2= 3When a=b(a^2-ab+b^2)/(a^2+ab+b^2)=[(b)^2-(b)b+b^2]/[(b)^2+(b)b+b^2]=(b^2-b^2+b^2)/(b^2+b^2+b^2)= b^2/3b^2= 1/3Option A is correct.
Question – 2
2. A task was to be completed in 20 days by a team. 12 members of the team were dropped and the remaining completed the work in 32 days.?
| Column-A | Column-B |
|---|---|
| Half of the number of members initially in the team | Number of members dropped from the team |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:A
- Answer Explanation:Let there be x members in the team initiallyx members complete the work in 20 daysWork done by 1 person in 1 day = 1/20xWork done by 1 person in 32 days = 32/20x = 8/5xNumber of persons after 12 members were dropped = (x-12)Work done by (x-12) persons in 32 days = work done by 1 person in 32days*number of persons1=(8/5x)*(x-12)5x=8x-8*125x=8x-968x-5x=963x=96x=32There were 32 persons initiallyHalf of 32=32/2=16Number of members dropped from the team = 12Option A is correct.
Question – 3
3. (a/b)^(x-6) = (b/a)^(2x-3)?
| Column-A | Column-B |
|---|---|
| -x^2. | x |
(Note: [a^2=a*a])
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:B
- Answer Explanation:(a/b)^(x-6) = (a/b)^[-(2x-3)](a/b)^(x-6) = (a/b)^(3-2x)x-6=3-2x2x+x=3+6x=9/3=3-x^2=-3^2=-9<3-x^2
Question – 4
4. The average of x and y is 25. x is 200% of the average of 5, 22, 13 and y.?
| Column-A | Column-B |
|---|---|
| x | (x+y)/3 |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:A
- Answer Explanation:The average of the numbers is (5+22+13+y)/4= (40+y)/4200% of this average is xx = 200/100*(40+y)/4x = (40+y)/22x = 40+y2x-y = 40 …….(1)Also the average of x and y is 25.(x+y)/2=25x+y = 25*2=50 ….(2)Add (1) and (2)2x-y+x+y=40+50=903x=90x=30Putting x=30 in x+y=50, we gety=50-30=20x=30(x+y)/3=50/3=16.67x>(x+y)/3Option A is correct.
Question – 5
5. x+1/y=1 and y+1/z=1.?
| Column-A | Column-B |
|---|---|
| z1/x | y1/z |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:C
- Answer Explanation:x+1/y=1x=1-1/y=(y-1)/y1/x=y/(y-1) …..(1)y+1/z=11/z=1-yz=1/(1-y) …..(2)Substituting values from (1) and (2), we getz+1/x = 1/(1-y)+y/(y-1)= 1/(1-y) – y/(1-y)= (1-y)/(1-y) = 1z+1/x=y+1/zOption C is correct.
Question – 6
6. Consider a regular polygon.?
| Column-A | Column-B |
|---|---|
| Number of diagonals | Number of interior angles |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:D
- Answer Explanation:PConsider a triangle:Number of diagonals = 0, number of interior angles = 3Consider a quadrilateral:Number of diagonals = 2, number of interior angles = 4Consider a pentagon:Number of diagonals = 5, number of interior angles = 5Hence, a relationship cannot be established.Option D is correct.
Question – 7
7. A pipe can fill a tank with water in 6 hours. Due to a leak, the tank is filled in 7 hours.?
| Column-A | Column-B |
|---|---|
| Time taken to empty a full tank through the leak | Two days |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:B
- Answer Explanation:Let the time taken to empty the tank due to the leak be x hours.Tank filled in one hour = 1/6Tank emptied in 1 hour = 1/x1/7=1/6-1/x6x=7x-427x-6x=42x=42Time taken to empty the tank through the leak = 42 hoursTwo days = 24*2=48 hoursOption B is correct
Question – 8
8. log[(x+y)/5] = 1/2 (logx+logy).?
| Column-A | Column-B |
|---|---|
| x/y | y/x |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:D
- Answer Explanation:log[(x+y)/5] = 1/2 (logx+logy)(x+y)/5 = sqrt(xy)(x+y)=5*sqrt(xy)Squaring both sides(x+y)^2 = 5^2*xyx^2+y^2+2xy = 25xyx^2+y^2=25xy-2xy = 23xyDividing both sides by xyx/y+y/x=23We cannot determine which of x/y and y/x is greater.Option D is correct.
Question – 9
9. A person covers half of his journey at 30 km/hr and the other half at 20 km/hr.?
| Column-A | Column-B |
|---|---|
| His average speed for the whole journey | His average speed for the first half of his journey |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:B
- Answer Explanation:Let the distance be 2x kmTime = Distance/speedTime for first half of journey = x/30Time for second half of journey = x/20Total time = x/30+x/20 = (20x+30x)/(20*30)= 50x/(20*30)= x/12Average speed = total distance/total time= (x+x)/(x/12)= 2x*12/x= 24 km/hrAverage speed for the first half of his journey = 30 km/hrOption B is correct.
Question – 10
10.
| Column-A | Column-B |
|---|---|
| The number of chords of a circle formed by joining the vertices of a cyclic pentagon. | The number of diameters of a circle formed by joining |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:A
- Answer Explanation:Any two points of the pentagon can be joined to form a chord.The number of points is given by C(5,2)C(5,2) = 5!/(2!3!)= 5*4/2 = 1010 chords can be drawn by joining the vertices of a cyclic pentagon.Of these 10 chords, 5 are the sides of the pentagon and 5 are the diagonals of the pentagon. None of these diagonals passes through the centre of the circle. Hence, the number of diameters is zero.Option A is true.
Question – 11
11. A stick of length 30cm is at a distance of 8 cm from the centre of a hemi-spherical urn.?
| Column-A | Column-B |
|---|---|
| Length of the longest stick that can be placed in the urn. | Diameter of the urn |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:C
- Answer Explanation:The longest stick that can be placed shall be equal to the diameter of the urn.Trivially option C is correct.
Question – 12
12. $20000 is lent out in two parts at 8% and 15% simple interest respectively. The net interest in a year is $2300.?
| Column-A | Column-B |
|---|---|
| First part | Second part |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:C
- Answer Explanation:Let the amount lent at 8% be $x and the amount lent at 15% be $(20000-x)Simple interest, SI=PRT/100, where P, R and T are principal, rate and time respectively.According to the conditions, we havex*8*1/100+(20000-x)*15*1/100=23008x+300000-15x=2300007x=70000x=$10000First part = $10000 and Second part = $20000-$10000=$10000Option C is correct.
Question – 13
13.
| Column-A | Column-B |
|---|---|
| The number of consecutive multiples of both 2 and 3 that lie between 200 and 350 | The number of consecutive multiples of both 2 and 3 that |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:C
- Answer Explanation:The numbers will be multiples of 6 since they are multiples of both 2 and 3.The numbers lie between 200 and 350 and form an A.P. with common difference 6.The first term of the A.P. will be 204 and the last term will be 348.The nth term of an AP with first term ‘a’ and common difference ‘d’ is given byTn=a+(n-1)d348 = 204 +(n-1)6144 = (n-1)6n = 24+1=2525 such numbers exist.The numbers will be multiples of 6 since they are multiples of both 2 and 3.The numbers lie between 350 and 500 and form an A.P. with common difference 6.The first term of the A.P. will be 354 and the last term will be 498.The nth term of an AP with first term ‘a’ and common difference ‘d’ is given byTn=a+(n-1)d498 = 354 +(n-1)6144 = (n-1)6n = 24+1=2525 such numbers exist.Option C is correct.
Question – 14
14. The difference between the areas of two concentric circles is 500*pi. The sum of the radii of the circles is equal to 50 cm.?
| Column-A | Column-B |
|---|---|
| Difference between the two radii | Radius of the smaller circle |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:B
- Answer Explanation:Let R and r be the two radii such that R>rArea of a circle = pi*r^2, where r is the radiuspi*R^2-pi*r^2=500*piR^2-r^2=500and R+r=50 cm(R^2-r^2) = (R+r)(R-r) = 500(50)(R-r)=500R-r=500/50 = 10 cmThe difference between the radii is 10 cm.R+r=50 and R-r=10Subtracting one relation from the other, we getR+r-R+r=50-10=402r=40r=20Radius of the smaller circle = 20cmOption B is correct.[pi=22/7][r^2=r*r]
Question – 15
15. In a school function, the girls have to raise one hand and the boys have to raise both their hands holding colored pieces of cloth in each. The boys hold three pieces, one green and two red and the girls hold one green piece each.The number of green cloth pieces is 120 and the number of red cloth pieces is 40.?
| Column-A | Column-B |
|---|---|
| Half the number of girls among the group | Twice the number of boys in the group |
- A. If the quantity on the left is greater
- B. If the quantity on the right is greater
- C. If both are equal
- D. If the relationship cannot be determined without further information
- Answer:A
- Answer Explanation:Let the number of boys be x and the number of girls be y.According to the conditions, we havex+y = 120 …(1)2x=40 …(2)x = 40/2=20y = 120 – x= 120 – 20 = 100Number of girls = 100Number of boys = 20Half the number of girls = 100/2=50Twice the number of boys = 20*2=40Option A is correct.