NIMCET Mathematics Previous Year Questions (PYQs) – Page 91 of 95
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The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls
combined is 50. The percentage of boys in the class is
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Solution
Given averages: Boys = 52, Girls = 42, Combined = 50.
Let the proportion of boys be \(p\) (so girls = \(1-p\)). Weighted mean gives: $$52p + 42(1-p) = 50$$ $$52p + 42 - 42p = 50 \Rightarrow 10p = 8 \Rightarrow p = 0.8.$$
Percentage of boys \(= 0.8 \times 100\% = \boxed{80\%}\).
The value of $\int ^{\pi/3}_{-\pi/3}\frac{x\sin x}{{\cos }^2x}dx$ is
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Solution
There are $4$ books on fairy tales, $5$ novels and $3$ plays. In how many ways can they be arranged in the order books on fairy tales, novels and then plays so that books of same category are put together
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Solution
Given:
$4$ books on fairy tales, $5$ novels, $3$ plays
The order of arrangement is fixed as:
Fairy Tales $\rightarrow$ Novels $\rightarrow$ Plays
Step 1: Arrange books within each category
Fairy tales books can be arranged among themselves $= 4!= 24$ ways
Novels can be arranged among themselves $= 5! = 120$ ways
Plays can be arranged among themselves $= 3! = 6$ ways
Step 2: Total number of arrangements
$= 4! \times 5! \times 3!$
$= 24 \times 120 \times 6$
$\therefore \boxed{= 17280 \text{ ways}}$
$4$ books on fairy tales, $5$ novels, $3$ plays
The order of arrangement is fixed as:
Fairy Tales $\rightarrow$ Novels $\rightarrow$ Plays
Step 1: Arrange books within each category
Fairy tales books can be arranged among themselves $= 4!= 24$ ways
Novels can be arranged among themselves $= 5! = 120$ ways
Plays can be arranged among themselves $= 3! = 6$ ways
Step 2: Total number of arrangements
$= 4! \times 5! \times 3!$
$= 24 \times 120 \times 6$
$\therefore \boxed{= 17280 \text{ ways}}$
If a, b, c are three non-zero vectors with no two of
which are collinear, a + 2b is collinear with c and b + 3c is collinear with a , then | a + 2b + 6c | will
be equal to
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Solution
If the angles of a triangle are in the ratio 2 : 3 : 7, then the ratio of the sides opposite to these
angles is
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Solution
How many even integers between 4000 and 7000 have four different digits?
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Solution
1: Thousands digit
The number is 4-digit and must be between 4000 and 7000.
So the thousands digit can be: 4, 5, or 6.
Step 2: Units digit
The number must be even, so the units digit must be one of {0, 2, 4, 6, 8}.
3: Count cases
- Case 1: Thousands digit = 4
Units digit ≠ 4 → possible units = {0, 2, 6, 8} → 4 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 4 × 8 × 7 = 224. - Case 2: Thousands digit = 5
Units digit can be {0, 2, 4, 6, 8} → 5 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 5 × 8 × 7 = 280. - Case 3: Thousands digit = 6
Units digit ≠ 6 → possible units = {0, 2, 4, 8} → 4 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 4 × 8 × 7 = 224.
4: Add results
Total = 224 + 280 + 224 = 728.
Final Answer:
There are 728 such even integers.
The equation of the tangent at any point of curve $x=a cos2t, y=2\sqrt{2} a sint$ with $m$ as its slope is
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Solution
Suppose a population $A$ has $100$ observations $101,102,\ldots,200$ and another population $B$ has $100$ observations $151,152,\ldots,250$. If $V_A$ and $V_B$ represent variance of the two populations respectively, then $\dfrac{V_A}{V_B}$ is
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Solution
Population $A$: $101, 102, \ldots, 200$
Population $B$: $151, 152, \ldots, 250$
Using Property:
Variance is independent of change of origin
i.e., if each observation is increased or decreased by a constant, variance remains unchanged
Here, each observation of $B$ = each observation of $A$ $+ 50$
$\Rightarrow$ Population $B$ is obtained by adding $50$ to each observation of population $A$
$\Rightarrow V_B = V_A$
$\therefore \dfrac{V_A}{V_B} = \boxed{1}$
Population $B$: $151, 152, \ldots, 250$
Using Property:
Variance is independent of change of origin
i.e., if each observation is increased or decreased by a constant, variance remains unchanged
Here, each observation of $B$ = each observation of $A$ $+ 50$
$\Rightarrow$ Population $B$ is obtained by adding $50$ to each observation of population $A$
$\Rightarrow V_B = V_A$
$\therefore \dfrac{V_A}{V_B} = \boxed{1}$
The arithmetic mean of two numbers a and b is 5, and the harmonic mean is 3.2. Find the numbers a and b.
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Solution
AM $=5$
$\frac{a+b}{2}=5$
$a+b=10$
HM $=3.2$
$\frac{2ab}{a+b}=3.2$
$\frac{2ab}{10}=3.2$
$ab=16$
Therefore, $a$ and $b$ are roots of
$x^2-10x+16=0$
$(x-8)(x-2)=0$
$\boxed{a=8,\quad b=2}$
$\frac{a+b}{2}=5$
$a+b=10$
HM $=3.2$
$\frac{2ab}{a+b}=3.2$
$\frac{2ab}{10}=3.2$
$ab=16$
Therefore, $a$ and $b$ are roots of
$x^2-10x+16=0$
$(x-8)(x-2)=0$
$\boxed{a=8,\quad b=2}$
Vertices of the vectors i - 2j + 2k , 2i + j - k and 3i - j + 2k form a triangle. This triangle is
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Solution
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