Free Quant Practice Test - CAT 

Thus the answer is $2^6=64$ cubes in the cup with the most sugar.

So the cups must contain 1, 2, 4, 8, 16, 32, and 64 cubes of sugar, with 64 being the most number of cubes in one cup.

$8^f=9\to 8=9^{\frac{1}{f}}$
$7^e=8\to 7^e=9^{\frac{1}{f}}\to 7=9^{\frac{1}{ef}}$
Continuing in the same way we get $3=9^{\frac{1}{abcdef}}$
$⟹3=3^{\frac{2}{abcdef}}$
$⟹\frac{2}{abcdef}=1\Rightarrow abcdef=2$

Approach 1
The expected value of the ratio of Frank's to Joe's score is 3:7, so Frank is
expected to win 3 games for each of Joe's 7. Extending the same ratio, 3:7 = 9:21. Thus the expected number of games in a match is 30.

Approach 2- Reverse Gear Approach
The ratios imply that Frank wins 7 out of 10 matches and Joe wins 3 out of 10 matches. Total number of matches = $\frac{7}{10}$xN + $\frac{3}{10}$xN (N is the number of matches possible)

Look for a multiple of 10 in the answer options. You can mark option (a) directly

The Euler's number $\varphi \left(n\right)$ for 240 is given as:

21, 40, 59, 78, 97, . . . , 4562 are in Arithmetic Progression with common difference as 19.
Number of terms $=\left[\frac{(4562-21)}{19}\right]+1$

$=\left(\frac{4541}{19}\right)+1=239+1=240$

Now, applying the interesting property of numbers:
"If a is prime to n, the number of terms of the Arithmetic Progression x, x+a, x+2a, . . . , x+(n-1)a which are prime to n is $\varphi \left(n\right)$.
$\varphi \left(n\right)$ = no. of positive integers smaller than n and co-prime to n.

Hence the number of positive integers in the above list whose HCF with 240 is not greater than 1
= The number of positive integers in the list which are prime to $240=\varphi \left(240\right)$ = 64.

Go from answer options

If x =1.41 =$\sqrt{2}$ then we get
 ${\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{\sqrt{2}}}}}=4$
Hence the equation is satisfied.

The figure looks as above.$\Delta$ABE and $\Delta$CDE are similar.
The area of each part is equal.
Thus the ratios of the areas= square of the ratios of the corresponding sides

$\Rightarrow$ If area of ∆CDE = x then area of $\Delta$ABE =9x

$\therefore$$\frac{CD}{AB}=\sqrt{\frac{9x}{x}}$ = 3:1

The digits of N can be 2, 4, 6 and 8. Since N is divisible by 4, the last two digits must be 24 or 28 or 44 or 48 or 64 or 68 or 84 or 88.
When the last two digits are the same, i.e., 44 or 88, there can be no more repetition; then, the first three digits can be chosen in 3! Ways. Thus, the number of possibilities = 3! x 2 = 12.
When the last two digits are distinct, we get the following cases:
i] Third digit is same as fourth digit.
Number of ways remaining two digits can be chosen = 2!
Ii] Third digit is same as second digit.
Number of ways = 2!
iii] Second digit is same as first digit.
Number of ways = 2!
Thus, when the number ends with 24, the number of possibilities = 2! + 2! + 2! = 6.
Since there are 6 cases where the last two digits are distinct, the total number of ways =

6 $\times$ 6+ 12 = 36 + 12 = 48. Hence, [b].

Check for values which are included in the range in 2 options and not there in two options i.e. substitute x=0, equation is not satisfied at x=0; thus option (a) and option (c) can be eliminated next, check for values which are there in option (b) and not there in option (d) or vice versa. Put x=1The equation is not satisfied at x=1, LHS= 6 and RHS=1. Thus the correct range is option (d)

Let x,y,x be the ages of 3 persons which lies between 60 and 65

The equation governing the problem $\frac{55n-41-47+(x+y+z)}{n+1}=56$

Also given n is a multiple of 7

55n - 88 +k = 56n + 56  where k=x+y+z, n+ 144=k  

Now k ranges between 3 $\times$ 61 and 3$\times$ 64

183 $\le$ n+44 $\le$ 192
39 $\le$ n $\le$ 48

Thus n =42, the latest number n + 1 =42+1=43

Shortcut

Go from answer options. The initial number of people in the group is a multiple of 7. The answer options give the number of initial people + 1. Thus the correct answer is option (b).

Each equal side of the triangle -

$=\frac{36-10}{2}=13cm$
$\therefore$ Height of the triangle
$={\sqrt{13}}^2-(\frac{10}{2}{)}^2$.
$=\sqrt{169-25}=12cm$
$\therefore$ Area of the triangle
$\frac{1}{2}\times 10\times 12=60c{m}^2$.

We know that the area of all triangles between two parallel lines and having a common base is the same

Thus, the triangle can also be represented as follows

we can say that, area (RQS) = area (RR1S)
Using the concept of graphical division, we can divide the figure into 8 equal regions as follows. 

The required region is $\frac{A}{8}$ units. 

Thus the remainder when divided by 100 is the last two digits = 50 itself

$\therefore x=\frac{45-3y-4z}{2}$

Also,$z<\frac{45-3y-4z}{2}<y$

Or, 6z+3y $<$ 45 and 5y+4z $>45.$

Possible values of y and z that satisfy the above equation is (7, 3); (9, 2) and (11, 1) in that order. Therefore, there are 3 different possible combinations.

abc leaves remainder of 1 when divided by $10⟹c=1$.
bca leaves remainder of 1 on dividing by $8⟹4b+2c+a=8m+1$ and similarly a+b+c = 3n+1.
On solving we get 3b = 8m -1 - 3n $⟹$ m = 2, 5, 8,... or 3p-1 format.
Now, 4b+2+a = 8m+1 $⟹$ 4b+a = 8m-1
For p=3, m=8, 4b+a = 63 means b > 8. So, p cannot be greater than 2.
For p=2, m=5, 4b+a = 39, if a = 3, 7 means b$\ge$8 which is not possible.
So, p=1, m=2, 4b+a = 15, possible value of a = 3.
Hence, choice (a) is the right option.

Let the area of the square= s, the area of the circle= c, and the area of the overlapping portion= x. The area of the circle not overlapped by the square is "c -x” and the area of the square not overlapped by the circle is "s - x”, so the difference between these two is (c -x) -(s -x) = c -s = 9${\pi }^2$ -4 .(approximately = 24.26). Hence the correct answer is option (b).

Let $u=\frac{a}{b}.$ Then the problem is equivalent to finding all possible rational numbers 'u' such that
$u+\frac{14}{9u}=k$, for some integer k.

This equation is equivalent to $9u^2-9uk+14=0$
$9u^2-9uk+14=0$ whose solutions are,

$u=\frac{9k\pm \sqrt{81k^2-504}}{18}$
$=\frac{k}{2}\pm \frac{1}{6}\sqrt{9k^2-56}$

Hence u is rational if and only if:
$\sqrt{9k^2-56}$ is rational, which is true if and only if $9k^2-56$ is a perfect square.
Suppose that $9k^2-56=S^2$ for some positive integer S.
(3k+S)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56.
So, (3k-S) and (3k+S) is one of the ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no integral solutions.
The cases (2, 28) and (4, 14) yield k = 5 and k = 3 respectively.

If k=5, then $u=\frac{1}{3}$ or $u=\frac{14}{3}.$
If k=3, then $u=\frac{2}{3}$ or $u=\frac{7}{3}.$
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3).

Let the number of correct answers be 'x', number of wrong answers be 'y' and number of questions not attempted be 'z'. 

Thus,x+y+z=50 .............(i)

And $x-\frac{y}{3}-\frac{z}{6}=32$

The second equation can be written as,
6x-2y-z=192 ...................(ii)
Adding the two equations we get,
$7x-y=242\Rightarrow x=\frac{242}{7}+\frac{y}{7}=\frac{242+y}{7}$

Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.
Hence option (b)

$\begin{array}{c}1\\ 2& 3\\ 4& 5& 6\end{array}$
This equation is based on the $\frac{n(n+1)}{2}$ pattern, where the last coin in the nth box will have the value $\frac{n(n+1)}{2}$.
Thus the last coin in the ${15}^{th}$ box will have a value $=\frac{15\times 16}{2}=120$
The first coin will have a value = 106
Thus the sum $=\frac{15}{2}(106+120)=1695.$

As there are 15 coins in the 15th box and it is 15 numbers ... it has to be a multiple of 15. Only option (b) satisfies this condition.

Use alligation,

Considering only the blue pigment in all the paints

Thus, 5 gms of each paint is used. Weight of red paint = $\frac{7}{10}\times 5$ =3.5 gms. Option (b)

Let's fix number 8 at a slot at the table, automatically, number 3 also gets fixed. Out of the remaining 6 delegates, 4 and 5 will be together and can be arranged in 2! Ways amongst each other. Also for each of these ways, number 1 can sit next to 4 in only one way.

Now 1,4,5 can be placed only in 2 ways(See below).

This question now becomes the arrangement of the remaining 3 people $\times$ 4= 3! $\times$ 4. Option (c)

$S_1=2$
$S_2=3$
$S_1+S_2=5$
Here n=2
Go from answer options
n(n+1)=2.3 =6
$\frac{1}{2}{n}^2=\frac{1}{2}\times 4=2$

$\frac{1}{2}n(n+3)=\frac{1}{2}\left(2\right)\left(5\right)=5$
n=2
$n^2=4$

Any number ending with 25 raised to any power will always end in 25.

 So, $\angle$BAC can't be less than 30 degrees so 'a' is the answer.

In a 45-45-90 triangle, sides are in the ratio 1:1:$\sqrt{2}$
Thus OB= 150 

In a 30-60-90 triangle, sides are in the ration 1:$\sqrt{3}$:2
Thus OA= $\frac{150}{\sqrt{3}}$
Distance travelled in 2 mins = AB
                                 = 150- 50$\sqrt{3}$ = 63.4 m
                                 speed = $\left(\frac{63.4}{2}\right)\times \left(\frac{60}{1000}\right)$= 1.9 kmph

f(98) = 98 + 2 = 100. Hence option (d)

$1000=18\times 55+10$
Problem now changes to $\frac{{15}^{10}}{19}$.
$\frac{15}{19}$ gives remainder -4. So the remainder of $\frac{(-4{)}^{10}}{15}$ or $\frac{(4{)}^{10}}{15}$ needs to be found.
Now $\frac{(4{)}^2}{15}$ gives a remainder of 1.
Going by the frequency method,
$\frac{(4{)}^{10}}{15}$ will give a remainder of $\frac{(1{)}^5}{15}$, i.e 1.